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周赛作业, Underfail(费用流)

开发技术 开发技术 2022-10-28 次浏览

题意是给你一个字符串,每个位置最多只能match k次,然后给你一些字符串,每个字符串有一些权值,每次可以匹配一个子段,匹配完之后子段使用次数 - 1, 分数 += wi

问满足条件下的最大权值

这个问题让我想起来了HDU的某题,好像20年的时候写过,就是区间最大覆盖问题,然后套板子,t了,不懂,遂去查题解

发现图是长这个样子

周赛作业, Underfail(费用流)

 

 所以需要加的边是l -> R + 1,要不然会出现负环,就会g

那就改变连边,然后字符串匹配可以用字符串hash加速到O(n)

就可以了

AC代码:

#include <bits/stdc++.h>
using namespace std;
constexpr int limit =  (100000 + 5);//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);
#define ff(a) printf("%dn",a );
#define pi(a, b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\Users\tiany\CLionProjects\akioi\data.txt", "rt", stdin)
#define FOUT freopen("C:\Users\tiany\CLionProjects\akioi\dabiao.txt", "wt", stdout)
typedef long long ll;
typedef unsigned long long ull;
char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf;

inline ll read() {
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    ll sign = 1, x = 0;
    char s = getchar();
    while (s > '9' || s < '0') {
        if (s == '-')sign = -1;
        s = getchar();
    }
    while (s >= '0' && s <= '9') {
        ll  x = (x << 3) + (x << 1) + s - '0';
        s = getchar();
    }
    return x * sign;
#undef getchar
}//快读
void print(ll x) {
    if (x / 10) print(x / 10);
    *O++ = x % 10 + '0';
}

void write(ll x, char c = 't') {
    if (x < 0)putchar('-'), x = -x;
    print(x);
    if (!isalpha(c))*O++ = c;
    fwrite(obuf, O - obuf, 1, stdout);
    O = obuf;
}

int n, m, k,vs,ve;
int head[limit],cnt;
int dist[limit],fa[limit],vis[limit],pre[limit];
struct node{
    int to, next, flow, w;
}edge[limit<<1];
void add_one(int u, int v, int flow, int w){
    edge[cnt].to = v;
    edge[cnt].flow = flow;
    edge[cnt].w = w;
    edge[cnt].next = head[u];
    head[u] = cnt++;
}
void add(int u, int v, int flow ,int w){
    add_one(u,v,flow,w);
    add_one(v,u,0,-w);
}
void init(int flag = 0){
    if(flag){
        memset(dist, INF, sizeof(dist));
        memset(vis, 0 , sizeof(vis));
    }else{
        memset(head, -1, sizeof(head));
        cnt = 0;
    }
}
int spfa(){
    queue<int>q;
    init(1);
    dist[vs] = 0;
    vis[vs] = 1;
    q.push(vs);
    pre[ve] = -1;
    while (q.size()){
        int u = q.front();
        q.pop();
        vis[u] = 0;
        for(int i = head[u]; ~i; i = edge[i].next){
            int v = edge[i].to, w = edge[i].w;
            int stream = edge[i].flow;
            if(dist[u] + w < dist[v] && stream > 0){
                dist[v] = dist[u] + w;
                pre[v] = i;
                fa[v] = u;
                if(!vis[v]){
                    vis[v] = 1;
                    q.push(v);
                }
            }
        }
    }
    return ~pre[ve];
}

int max_flow,min_cost;
void dinic(){
    while (spfa()){
        int min_flow = INF;
        for(int i = ve; i != vs; i = fa[i]){
            min_flow = min(min_flow, edge[pre[i]].flow);
        }
        max_flow += min_flow;
        min_cost  += dist[ve] * min_flow;
        for(int i = ve; i != vs; i = fa[i]){
            edge[pre[i]].flow -= min_flow;
            edge[pre[i] ^ 1].flow += min_flow;
        }
    }
}



struct hasher{
    ll hash[limit];
    ll p[limit];
    void init(string & str){
        p[0] = 1;
        for(int i = 1; i <= n; ++i){
            p[i] = p[i - 1] * 26;
            hash[i] = hash[i - 1] * 26 + str[i];
        }
    }
    ll get(int l, int r){
        return hash[r] - hash[l - 1] * p[r - l + 1];
    }
    ll calc(string & str){
        ll ans = 0;
        for(auto && it : str){
            ans = ans * 26 + it;
        }
        return ans;
    }
};
int a[limit];
int l[limit], r[limit];
int c[limit];
int rev[limit];
void solve(){
    init();
    cin>>n;
    string str;
    cin>>str;
    str = " " + str;
    vs = 80001, ve = vs + 1;
    int tot = 0;
    int tot1 = 0;
    hasher h;
    h.init(str);
    cin>>k;
    rep(i,1,k){
        string s;
        ll w;
        cin>>s>>w;
        ll val = h.calc(s);
        rep(hh,1, n){ //暴力匹配
            if(hh + s.length() - 1 > n)break;
            if(h.get(hh, hh + s.length() - 1) == val){
                a[++tot1] = hh;
                a[++tot1] = hh + s.length();
                l[++tot] = hh;
                r[tot] = hh + s.length();
                c[tot] = w;
            }
        }
    }
    cin>>k;
    sort(a + 1, a + 1 + tot1);
    tot1 = unique(a + 1, a + 1 + tot1) - a - 1;
    rep(i,1,tot1){
        rev[a[i]] = i; //反射
    }
    rep(i,1,tot1){
        add(i, i + 1, k, 0);
    }
    add(vs, 1, k, 0);
    add(tot1 + 1, ve, k, 0);
    rep(i,1,tot){
        add(rev[l[i]], rev[r[i]], 1, -c[i]);
    }
    dinic();
    cout<<-min_cost<<endl;

};
int32_t main() {
#ifdef LOCAL
    FOPEN;
//    FOUT;
#endif
    FASTIO
//    int kase;
//    cin>>kase;
//    while (kase--)
    solve();
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "sn";
    return 0;
}

 

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