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一些最短路的题目

互联网 diligentman 2周前 (05-03) 8次浏览

HDU1596
Problem Description
XX星球有很多城市,每个城市之间有一条或多条飞行通道,但是并不是所有的路都是很安全的,每一条路有一个安全系数s,s是在 0 和 1 间的实数(包括0,1),一条从u 到 v 的通道P 的安全度为Safe§ = s(e1)*s(e2)…*s(ek) e1,e2,ek是P 上的边 ,现在8600 想出去旅游,面对这这么多的路,他想找一条最安全的路。但是8600 的数学不好,想请你帮忙 _

Input
输入包括多个测试实例,每个实例包括:
第一行:n。n表示城市的个数n<=1000;
接着是一个n*n的矩阵表示两个城市之间的安全系数,(0可以理解为那两个城市之间没有直接的通道)
接着是Q个8600要旅游的路线,每行有两个数字,表示8600所在的城市和要去的城市

Output
如果86无法达到他的目的地,输出"What a pity!",
其他的输出这两个城市之间的最安全道路的安全系数,保留三位小数。

Sample Input
3
1 0.5 0.5
0.5 1 0.4
0.5 0.4 1
3
1 2
2 3
1 3

Sample Output
0.500
0.400
0.500

floyd算法

#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdlib>

using namespace std;
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define inf 0x3f3f3f3f
#define ll long long
int n;
double ave[1005][1005];
void init()
{
    for(int i = 1;i <= n; i++)
        for(int j = 1; j <= n; j++)
        scanf("%lf",&ave[i][j]);
}
void floyd()
{
    for(int k = 1;k <= n;k++){
        for(int i = 1; i <= n ;i++){
            for( int j = 1;j <= n; j++)
            ave[i][j] = max(ave[i][j],ave[i][k]*ave[k][j]);
        }
    }
}
int main()
{   int q,a,b;
    while(~scanf("%d",&n)){
        init();
        floyd();
        scanf("%d",&q);
        for(int i = 0 ;i < q;i++){
            scanf("%d%d",&a,&b);
            if(ave[a][b] == 0) cout<<"What a pity!"<<endl;
            else printf("%.3lfn",ave[a][b]);
        }
    }
return 0;
}

HDU1874
Problem Description
某省自从实行了很多年的畅通工程计划后,终于修建了很多路。不过路多了也不好,每次要从一个城镇到另一个城镇时,都有许多种道路方案可以选择,而某些方案要比另一些方案行走的距离要短很多。这让行人很困扰。

现在,已知起点和终点,请你计算出要从起点到终点,最短需要行走多少距离。

Input
本题目包含多组数据,请处理到文件结束。
每组数据第一行包含两个正整数N和M(0<N<200,0<M<1000),分别代表现有城镇的数目和已修建的道路的数目。城镇分别以0~N-1编号。
接下来是M行道路信息。每一行有三个整数A,B,X(0<=A,B<N,A!=B,0<X<10000),表示城镇A和城镇B之间有一条长度为X的双向道路。
再接下一行有两个整数S,T(0<=S,T<N),分别代表起点和终点。

Output
对于每组数据,请在一行里输出最短需要行走的距离。如果不存在从S到T的路线,就输出-1.

Sample Input
3 3
0 1 1
0 2 3
1 2 1
0 2
3 1
0 1 1
1 2

Sample Output
2
-1

#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdlib>

using namespace std;
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define inf 0x3f3f3f3f
int n,m,ave[205][205];
void floyd()
{
    int k,i,j;
    for(k = 1; k <= n;k++){
        for(i = 1 ;i <=n ;i++){
            for(j = 1;j <= n; j++){
                if(ave[i][j]>ave[i][k]+ave[k][j])
                ave[i][j] = ave[i][k]+ave[k][j];
            }
        }
    }
}
int main()
{
    int a,b,c;
    while(cin>>n>>m){
        for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        {
            ave[i][j]= i==j?0:inf;
        }
        for(int i = 0 ;i < m ;i ++){
            cin>>a>>b>>c;
            a++;b++;
            if(ave[a][b]>c)
            ave[a][b] = ave[b][a] = c;
        }
        floyd();
        cin>>a>>b;
        a++;b++;
        if(ave[a][b] != inf)
        cout<<ave[a][b]<<endl;
        else cout<<-1<<endl;
    }
return 0;
}

HDU2112
Problem Description
经过锦囊相助,海东集团终于度过了危机,从此,HDU的发展就一直顺风顺水,到了2050年,集团已经相当规模了,据说进入了钱江肉丝经济开发区500强。这时候,XHD夫妇也退居了二线,并在风景秀美的诸暨市浬浦镇陶姚村买了个房子,开始安度晚年了。
这样住了一段时间,徐总对当地的交通还是不太了解。有时很郁闷,想去一个地方又不知道应该乘什么公交车,在什么地方转车,在什么地方下车(其实徐总自己有车,却一定要与民同乐,这就是徐总的性格)。
徐总经常会问蹩脚的英文问路:“Can you help me?”。看着他那迷茫而又无助的眼神,热心的你能帮帮他吗?
请帮助他用最短的时间到达目的地(假设每一路公交车都只在起点站和终点站停,而且随时都会开)。

Input
输入数据有多组,每组的第一行是公交车的总数N(0<=N<=10000);
第二行有徐总的所在地start,他的目的地end;
接着有n行,每行有站名s,站名e,以及从s到e的时间整数t(0<t<100)(每个地名是一个长度不超过30的字符串)。
note:一组数据中地名数不会超过150个。
如果N==-1,表示输入结束。

Output
如果徐总能到达目的地,输出最短的时间;否则,输出“-1”。

Sample Input
6
xiasha westlake
xiasha station 60
xiasha ShoppingCenterofHangZhou 30
station westlake 20
ShoppingCenterofHangZhou supermarket 10
xiasha supermarket 50
supermarket westlake 10
-1

Sample Output
50

Hint:
The best route is:
xiasha->ShoppingCenterofHangZhou->supermarket->westlake

虽然偶尔会迷路,但是因为有了你的帮助
从此还是过上了幸福的生活。

――全剧终――

#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<cstdlib>

using namespace std;
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define inf 0x3f3f3f3f
int n,cnt,ave[155][155],vis[155],dis[155];
map<string,int>p;
string s,t;

int main()
{   IOS;
    while(cin>>n&&n!=-1){
        cnt = 0;
        if(n == 0) {
            cout<<0<<endl;
            continue;
        }
        memset(ave,inf,sizeof(ave));
        p.clear();
        cin>>s>>t;
        p[s] = ++cnt;
        int start = p[s];
        if(!p[t])
        p[t] = ++cnt;
        int end = p[t];

        int c;
        for(int i = 1;i <= n;i++){
            cin>>s>>t>>c;
            if(!p[s]) p[s] = ++cnt;
            if(!p[t]) p[t] = ++cnt;
            if(ave[p[s]][p[t]] > c) ave[p[s]][p[t]] = ave[p[t]][p[s]] = c;
        }
        if(start == end) {
            cout<<0<<endl;
            continue;
        }
        for(int i = 1;i <= cnt;i++){
            if(i != start ) {
                dis[i] = ave[start][i];
                vis[i] = 1;
            }
            else {
                dis[i] = 0;
                vis[i] = 0;
            }
        }
        for(int i = 1;i < cnt;i++){
            int ans,k;
            ans = inf;k = 1;
            for(int j = 1;j <= cnt ;j++){
                if(vis[j] && dis[j] < ans) {
                    ans = dis[j];
                    k = j;
                }
            }
            vis[k] = 0;
            for(int j = 1;j <= cnt;j++){
                if(vis[j] && ans+ave[k][j] < dis[j]) dis[j] = ans + ave[k][j];
            }

        }
        if(dis[end] == inf) cout<<-1<<endl;
        else cout<<dis[end]<<endl;
    }
return 0;
}

HDU2680

Problem Description
One day , Kiki wants to visit one of her friends. As she is liable to carsickness , she wants to arrive at her friend’s home as soon as possible . Now give you a map of the city’s traffic route, and the stations which are near Kiki’s home so that she can take. You may suppose Kiki can change the bus at any station. Please find out the least time Kiki needs to spend. To make it easy, if the city have n bus stations ,the stations will been expressed as an integer 1,2,3…n.

Input
There are several test cases.
Each case begins with three integers n, m and s,(n<1000,m<20000,1=<s<=n) n stands for the number of bus stations in this city and m stands for the number of directed ways between bus stations .(Maybe there are several ways between two bus stations .) s stands for the bus station that near Kiki’s friend’s home.
Then follow m lines ,each line contains three integers p , q , t (0<t<=1000). means from station p to station q there is a way and it will costs t minutes .
Then a line with an integer w(0<w<n), means the number of stations Kiki can take at the beginning. Then follows w integers stands for these stations.

Output
The output contains one line for each data set : the least time Kiki needs to spend ,if it’s impossible to find such a route ,just output “-1”.

Sample Input
5 8 5
1 2 2
1 5 3
1 3 4
2 4 7
2 5 6
2 3 5
3 5 1
4 5 1
2
2 3
4 3 4
1 2 3
1 3 4
2 3 2
1
1

Sample Output
1
-1

#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<cstdlib>

using namespace std;
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define inf 0x3f3f3f3f
int n,m,s,ave[1010][1010],dis[1010],vis[1010];

int main()
{   int a,b,c;
    IOS;
    while(cin>>n>>m>>s){
        n++;
        memset(ave,inf,sizeof(ave));
        for(int i = 1;i <= m;i++){
            cin>>a>>b>>c;
            if(ave[a][b] > c) ave[a][b] = c;//注意是单向
        }
        int w;
        cin>>w;
        for(int i = 1;i <= w;i++){
            cin>>a;
            ave[n][a] = 0;
        }
        for(int i = 1;i <= n;i++){
            if(i != n) {
                dis[i] = ave[n][i];
                vis[i] = 1;
            }
            else {
                dis[i] = 0;
                vis[i] = 0;
            }
        }
        for(int i = 1; i <= n; i++){
            int k,ans;
            ans = inf;
            for(int j = 1;j <= n;j++){
                if(vis[j] && dis[j] < ans) {
                    ans = dis[j];
                    k = j;
                }
            }
            if(ans == inf) break;
            vis[k] = 0;
            for(int j = 1;j <= n;j++){
                if(vis[j] && ans + ave[k][j] < dis[j]) dis[j] = ans + ave[k][j];
            }
        }
        if(dis[s] != inf)
        cout<<dis[s]<<endl;
        else cout<<-1<<endl;
    }
return 0;
}


HDU1690
Problem Description
Because of the huge population of China, public transportation is very important. Bus is an important transportation method in traditional public transportation system. And it’s still playing an important role even now.
The bus system of City X is quite strange. Unlike other city’s system, the cost of ticket is calculated based on the distance between the two stations. Here is a list which describes the relationship between the distance and the cost.

Your neighbor is a person who is a really miser. He asked you to help him to calculate the minimum cost between the two stations he listed. Can you solve this problem for him?
To simplify this problem, you can assume that all the stations are located on a straight line. We use x-coordinates to describe the stations’ positions.

Input
The input consists of several test cases. There is a single number above all, the number of cases. There are no more than 20 cases.
Each case contains eight integers on the first line, which are L1, L2, L3, L4, C1, C2, C3, C4, each number is non-negative and not larger than 1,000,000,000. You can also assume that L1<=L2<=L3<=L4.
Two integers, n and m, are given next, representing the number of the stations and questions. Each of the next n lines contains one integer, representing the x-coordinate of the ith station. Each of the next m lines contains two integers, representing the start point and the destination.
In all of the questions, the start point will be different from the destination.
For each case,2<=N<=100,0<=M<=500, each x-coordinate is between -1,000,000,000 and 1,000,000,000, and no two x-coordinates will have the same value.

Output
For each question, if the two stations are attainable, print the minimum cost between them. Otherwise, print “Station X and station Y are not attainable.” Use the format in the sample.

Sample Input
2
1 2 3 4 1 3 5 7
4 2
1
2
3
4
1 4
4 1
1 2 3 4 1 3 5 7
4 1
1
2
3
10
1 4

Sample Output
Case 1:
The minimum cost between station 1 and station 4 is 3.
The minimum cost between station 4 and station 1 is 3.
Case 2:
Station 1 and station 4 are not attainable.

#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<cstdlib>

using namespace std;
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define inf 0x3f3f3f3f3f3f3f3f
#define ll long long
ll l1,l2,l3,l4,c1,c2,c3,c4,n,m,k;
ll ave[110][110],dis[110];

int main()
{   IOS;
    int nn,op;
    cin>>nn;
    op = 1;
    while(nn--){
        cin>>l1>>l2>>l3>>l4>>c1>>c2>>c3>>c4;
        cin>>n>>m;
        for(int i = 1;i <= n;i++){
            cin>>dis[i];
        }
        for(int i = 1;i <= n;i++){
            for(int j = i;j <= n;j++){
                k = abs(dis[i] - dis[j]);
                if(k == 0) ave[i][j] = ave[j][i] = 0;
                else if(k <= l1) ave[i][j] = ave[j][i] = c1;
                else if(k <= l2) ave[i][j] = ave[j][i] = c2;
                else if(k <= l3) ave[i][j] = ave[j][i] = c3;
                else if(k <= l4) ave[i][j] = ave[j][i] = c4;
                else ave[i][j] = ave[j][i] = inf;
            }
        }
        for(int k = 1;k <= n;k++)
            for(int i = 1;i <= n;i++)
                for(int j = 1;j <= n;j++)
                ave[i][j] = min(ave[i][j],ave[i][k] + ave[k][j]);
        printf("Case %d:n",op++);
        ll a,b;
        for(int i = 1;i <= m;i++){
            cin>>a>>b;
            if(ave[a][b] == inf)
            printf("Station %lld and station %lld are not attainable.n",a,b);
            else  printf("The minimum cost between station %lld and station %lld is %lld.n",a,b,ave[a][b]);
        }
    }
return 0;
}


POJ3268
Description

One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1…N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires Ti (1 ≤ Ti ≤ 100) units of time to traverse.

Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow’s return route might be different from her original route to the party since roads are one-way.

Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?

Input

Line 1: Three space-separated integers, respectively: N, M, and X
Lines 2…M+1: Line i+1 describes road i with three space-separated integers: Ai, Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti time units to traverse.
Output

Line 1: One integer: the maximum of time any one cow must walk.
Sample Input

4 8 2
1 2 4
1 3 2
1 4 7
2 1 1
2 3 5
3 1 2
3 4 4
4 2 3
Sample Output

10
Hint

Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units.
Source

#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdlib>

using namespace std;
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define inf 0x3f3f3f3f
#define ll long long
int n,m,x,ave[1005][1005],dis[1005],flag[1005];
void dij(int start)
{
    for(int i = 1; i <= n; i++){
        if(i != start){
            dis[i] = ave[start][i];
            flag[i] = 1;
        }
        else {
            dis[i] = flag[i] = 0;
        }
    }
    int num = 1;
    while(num < n){
        int ans,k;
        ans = inf;
        for(int i = 1;i <= n;i++){
            if(flag[i] && dis[i] < ans) {
                ans = dis[i];
                k = i;
            }
        }
        flag[k] = 0;
        for(int i = 1;i <= n;i++){
            if(flag[i] && dis[i] > dis[k] + ave[k][i]) dis[i] = dis[k] + ave[k][i];
        }
        num++;
    }
}
int main()
{   IOS;
    cin>>n>>m>>x;
    memset(ave,inf,sizeof(ave));
    for(int i = 1;i <= m;i++){
        int a,b,c;
        cin>>a>>b>>c;
        if(ave[a][b] > c) ave[a][b] = c;
    }
    dij(x);
    int ma[1005];
    for(int i = 1;i <= n;i++){
        ma[i] = dis[i];
        for(int j = i+1;j <= n;j++) swap(ave[i][j],ave[j][i]);
    }
    dij(x);
    int ans = 0;
    for(int i = 1;i <= n;i++){
        if(ma[i] + dis[i] > ans) ans = ma[i] + dis[i];
     }
     cout<<ans<<endl;
     return 0;
}


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