hdu 1269 迷宫城堡

迷宫城堡

Time Limit: 2000/1000 MS (Java/Others)    

Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 6826    Accepted Submission(s): 3034

Problem Description

为了训练小希的方向感，Gardon建立了一座大城堡，里面有N个房间(N<=10000)和M条通道(M<=100000)，每个通道都是单向的，就是说若称某通道连通了A房间和B房间，只说明可以通过这个通道由A房间到达B房间，但并不说明通过它可以由B房间到达A房间。Gardon需要请你写个程序确认一下是否任意两个房间都是相互连通的，即：对于任意的i和j，至少存在一条路径可以从房间i到房间j，也存在一条路径可以从房间j到房间i。

 

 

Input

输入包含多组数据，输入的第一行有两个数：N和M，接下来的M行每行有两个数a和b，表示了一条通道可以从A房间来到B房间。文件最后以两个0结束。

 

 

Output

对于输入的每组数据，如果任意两个房间都是相互连接的，输出"Yes"，否则输出"No"。

 

 

Sample Input

3 3 1 2 2 3 3 1 3 3 1 2 2 3 3 2 0 0

 

 

Sample Output

Yes

No

 

 

Author

Gardon

 

 

Source

HDU 2006-4 Programming Contest 

 

解题：强连通图判断，如果当前迷宫是一个强连通图，那么即Yes

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
const int maxv = 10010;
const int maxe = 100010;
int dfn[maxv],low[maxv],n,m,index,scc;
vector<int>g[maxv];
bool instack[maxv];
stack<int>s;
void tarjan(int u){
    s.push(u);
    instack[u] = true;
    dfn[u] = low[u] = ++index;
    for(int i = 0; i < g[u].size(); i++){
        int v = g[u][i];
        if(!dfn[v]){
            tarjan(v);
            low[u] = min(low[u],low[v]);
        }else if(instack[u] && low[u] > dfn[v]) low[u] = dfn[v];
    }
    if(dfn[u] == low[u]){
        int v;
        scc++;
        do{
            v = s.top();
            s.pop();
        }while(v != u);
    }
}
int main(){
    int i,j,u,v;
    while(scanf("%d%d",&n,&m),n+m){
        for(i = 1; i <= n; i++)
            g[i].clear();
        memset(dfn,0,sizeof(dfn));
        memset(low,0,sizeof(low));
        memset(instack,false,sizeof(instack));
        for(i = 0; i < m; i++){
            scanf("%d%d",&u,&v);
            g[u].push_back(v);
        }
        while(!s.empty()) s.pop();
        index = scc = 0;
        for(i = 1; i <= n; i++)
            if(!dfn[i]) tarjan(i);
        scc==1?puts("Yes"):puts("No");
    }
    return 0;
}

 

上面的代码有点傻逼，冗余了一些计算。。。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
const int maxv = 10010;
const int maxe = 100010;
int dfn[maxv],low[maxv],n,m,index,scc;
vector<int>g[maxv];
bool instack[maxv],flag;
stack<int>s;
void tarjan(int u){
    s.push(u);
    instack[u] = true;
    dfn[u] = low[u] = ++index;
    for(int i = 0; i < g[u].size(); i++){
        int v = g[u][i];
        if(!dfn[v]){
            tarjan(v);
            low[u] = min(low[u],low[v]);
        }else if(instack[u] && low[u] > dfn[v]) low[u] = dfn[v];
    }
    if(dfn[u] == low[u]){
        int v,o = 0;
        do{
            v = s.top();
            s.pop();
            o++;
        }while(v != u);
        if(o == n) flag = true;
    }
}
int main(){
    int i,j,u,v;
    while(scanf("%d%d",&n,&m),n+m){
        for(i = 1; i <= n; i++){
            g[i].clear();
            dfn[i] = 0;
            low[i] = 0;
            instack[i] = false;
        }
        for(i = 0; i < m; i++){
            scanf("%d%d",&u,&v);
            g[u].push_back(v);
        }
        while(!s.empty()) s.pop();
        flag = index = 0;
        tarjan(1);
        flag?puts("Yes"):puts("No");
    }
    return 0;
}

 

美观点

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
const int maxv = 10010;
const int maxe = 100010;
int dfn[maxv],low[maxv],n,m,index,scc;
vector<int>g[maxv];
bool instack[maxv];
stack<int>s;
bool tarjan(int u) {
    s.push(u);
    instack[u] = true;
    dfn[u] = low[u] = ++index;
    for(int i = 0; i < g[u].size(); i++) {
        int v = g[u][i];
        if(!dfn[v]) {
            if(tarjan(v)) return true;
            low[u] = min(low[u],low[v]);
        } else if(instack[u] && low[u] > dfn[v]) low[u] = dfn[v];
    }
    if(dfn[u] == low[u]) {
        int v,o = 0;
        do {
            v = s.top();
            s.pop();
            o++;
        } while(v != u);
        if(o == n) return true;
    }
    return false;
}
int main() {
    int i,j,u,v;
    while(scanf("%d%d",&n,&m),n+m) {
        for(i = 1; i <= n; i++) {
            g[i].clear();
            dfn[i] = 0;
            low[i] = 0;
            instack[i] = false;
        }
        for(i = 0; i < m; i++) {
            scanf("%d%d",&u,&v);
            g[u].push_back(v);
        }
        while(!s.empty()) s.pop();
        index = 0;
        tarjan(1)?puts("Yes"):puts("No");
    }
    return 0;
}