• 微信公众号:美女很有趣。 工作之余,放松一下,关注即送10G+美女照片!

题解 茅山道术

开发技术 开发技术 4小时前 2次浏览

传送门

先把相邻的同色的点缩点
每个点向上一个同色的点连边
发现就是求选出几条边,使它们不相交的方案数
考虑令 (dp[i][j]) 为考虑到位置 (i),上一次选点为 (j) 时的方案数
可以树状数组优化转移

Code:
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 1000010
#define ll long long
#define reg register int
//#define int long long 

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n;
int c[N];
const ll mod=1e9+7;
inline void md(ll& a, ll b) {a+=b; a=a>=mod?a-mod:a;}

namespace task1{
	int pre[N], lst[N], bkp[N], tot;
	ll dp[2010][2010];
	void solve() {
		dp[0][0]=1;
		memset(lst, -1, sizeof(lst));
		memcpy(bkp, c, sizeof(c));
		for (int pos2=1; pos2<=n; ++pos2)
			if (bkp[pos2-1]!=bkp[pos2]) c[++tot]=bkp[pos2];
		for (int i=1; i<=tot; ++i) {
			pre[i]=lst[c[i]];
			lst[c[i]]=i;
		}
		//cout<<"pre: "; for (int i=1; i<=tot; ++i) cout<<pre[i]<<' '; cout<<endl;
		for (int i=1; i<=tot; ++i) {
			for (int j=0; j<i; ++j) {
				dp[i][j] = dp[i-1][j];
				if (pre[i]>=j) dp[i][i]=(dp[i][i]+dp[i-1][j])%mod;
			}
		}
		ll ans=0;
		for (int i=0; i<=tot; ++i) ans=(ans+dp[tot][i])%mod;
		printf("%lldn", ans);
		exit(0);
	}
}

namespace task2{
	int pre[N], lst[N], bkp[N], tot;
	ll dp[N];
	inline void upd(int i, ll dat) {dat%=mod; for (; i<=tot+1; i+=i&-i) md(dp[i], dat);}
	inline ll query(int i) {ll ans=0; for (; i; i-=i&-i) md(ans, dp[i]); return ans;}
	void solve() {
		memset(lst, -1, sizeof(lst));
		memcpy(bkp, c, sizeof(c));
		for (int pos2=1; pos2<=n; ++pos2)
			if (bkp[pos2-1]!=bkp[pos2]) c[++tot]=bkp[pos2];
		for (int i=1; i<=tot; ++i) {
			pre[i]=lst[c[i]];
			lst[c[i]]=i;
		}
		upd(1, 1);
		//cout<<"pre: "; for (int i=1; i<=tot; ++i) cout<<pre[i]<<' '; cout<<endl;
		ll tem;
		for (int i=1; i<=tot; ++i) if (~pre[i]) {
			tem=query(pre[i]+1);
			if (tem) upd(i+1, tem);
		}
		printf("%lldn", query(tot+1));
		exit(0);
	}
}

signed main()
{
	freopen("magic.in", "r", stdin);
	freopen("magic.out", "w", stdout);
	
	n=read();
	for (int i=1; i<=n; ++i) c[i]=read();
	task2::solve();
	
	return 0;
}

程序员灯塔
转载请注明原文链接:题解 茅山道术
喜欢 (0)