BZOJ2208:[JSOI2010]连通数——题解

https://www.lydsy.com/JudgeOnline/problem.php?id=2208

 floyd压位是神马东西……

 我们tarjan缩点之后反向拓扑就可以记录联通块可达状态,然后可达就sz[i]*sz[j]就行了。

#include<map>
#include<cmath>
#include<stack>
#include<queue>
#include<cstdio>
#include<cctype>
#include<bitset>
#include<vector>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=2005;
struct edge{
    int cnt,head[N];
    int to[N*N],nxt[N*N];
    edge(){
        cnt=0;memset(head,0,sizeof(head));
    }
    inline void add(int u,int v){
        to[++cnt]=v;nxt[cnt]=head[u];head[u]=cnt;
    }
}e,f;
char s[N];
int sz[N],dfn[N],low[N],to[N],t,l;
bool inq[N];
stack<int>q;
void tarjan(int u){
    dfn[u]=low[u]=++t;
    q.push(u);inq[u]=1;
    for(int i=e.head[u];i;i=e.nxt[i]){
        int v=e.to[i];
        if(!dfn[v]){
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }else if(inq[v]){
            low[u]=min(low[u],dfn[v]);
        }
    }
    if(dfn[u]==low[u]){
        int v;l++;
        do{
            v=q.top();q.pop();inq[v]=0;
            to[v]=l;sz[l]++;
        }while(v!=u);
    }
}
int deg[N];
queue<int>que;
bitset<N>d[N];
void topu(int n){
    for(int i=1;i<=l;i++){
        d[i][i]=1;
        if(!deg[i])que.push(i);
    }
    while(!que.empty()){
        int u=que.front();que.pop();
        for(int i=f.head[u],v;i;i=f.nxt[i]){
            deg[v=f.to[i]]--;
            d[v]|=d[u];
            if(!deg[v])que.push(v);
        }
    }
}
int main(){
    int n;scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%s",s+1);
        for(int j=1;j<=n;j++)
            if(s[j]-'0')e.add(i,j);
    }
    for(int i=1;i<=n;i++)
        if(!dfn[i])tarjan(i);
    for(int u=1;u<=n;u++){
        for(int i=e.head[u];i;i=e.nxt[i]){
            int v=e.to[i];
            if(to[u]==to[v])continue;
            bool flag=1;
            for(int j=f.head[to[v]];j&&flag;j=f.nxt[j])
                if(f.to[j]==to[u])flag=0;
            if(flag)f.add(to[v],to[u]),deg[to[u]]++;
        }
    }
    topu(l);
    int ans=0;
    for(int i=1;i<=l;i++)
        for(int j=1;j<=l;j++)
            if(d[i][j])ans+=sz[i]*sz[j];
    printf("%d\n",ans);
    return 0;
}

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+本文作者:luyouqi233。               +

+欢迎访问我的博客:http://www.cnblogs.com/luyouqi233/ +

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posted @ 2018-06-21 19:43  luyouqi233  阅读(203)  评论(0编辑  收藏  举报