迷宫城堡
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9893 Accepted Submission(s): 4433
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
#include <cstdio>
#include <cstring>
#include <algorithm>
#define maxn 10000 + 100
#define maxm 100000 + 1000
using namespace std;
int n, m;
struct node {
int u, v, next;
};
node edge[maxm];
int head[maxn], cnt;
int low[maxn], dfn[maxn];
int dfs_clock;
int Stack[maxn];
bool Instack[maxn];
int top;
int Belong[maxn] , scc_clock;
void init(){
cnt = 0;
memset(head, -1, sizeof(head));
}
void addedge(int u, int v){
edge[cnt] = {u, v, head[u]};
head[u] = cnt++;
}
void getmap(){
while(m--){
int a, b;
scanf("%d%d", &a, &b);
addedge(a, b);
}
}
void tarjan(int u, int per){
int v;
low[u] = dfn[u] = ++dfs_clock;
Stack[top++] = u;
Instack[u] = true;
int have = 1;
for(int i = head[u]; i != -1; i = edge[i].next){
v = edge[i].v;
if(v == per && have){
have = 0;
continue;
}
if(!dfn[v]){
tarjan(v, u);
low[u] = min(low[v], low[u]);
}
else if(Instack[v]){
low[u] = min(low[u], dfn[v]);
}
}
if(dfn[u] == low[u]){
scc_clock++;
do{
v = Stack[--top];
Instack[v] = false;
Belong[v] = scc_clock;
}while(u != v);
}
}
void find(){
memset(low, 0, sizeof(low));
memset(dfn, 0, sizeof(dfn));
memset(Instack, false, sizeof(Instack));
memset(Belong, 0, sizeof(Belong));
dfs_clock = scc_clock = top = 0;
for(int i = 1; i <= n; ++i){
if(!dfn[i])
tarjan(i, i);
}
}
void solve(){
if(scc_clock == 1)
printf("Yes\n");
else
printf("No\n");
}
int main (){
while(scanf("%d%d", &n, &m), n || m){
init();
getmap();
find();
solve();
}
return 0;
}