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题意：给你n个点，编号为 0 ------ n - 1，然后下边每个点对应着m条边（注意点出现的顺序不是固定的），然后让你输出桥。

思路：本来就是一道tarjan求桥模板题啊，但是debug了1小时，求助才发现给的边并不一定相邻。可以给的点数应该不是很大，所以可以用矩阵标记加边。

AC Code：

#include<iostream>
#include<cstring>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<cmath>
#include<cstdio>
#include<iomanip>
#include<sstream>
#include<algorithm>using namespace std;
#define read(x) scanf("%d",&x)
#define Read(x,y) scanf("%d%d",&x,&y)
#define sRead(x,y,z)  scanf("%d%d%d",&x,&y,&z)
#define gc(x)  scanf(" %c",&x);
#define mmt(x,y)  memset(x,y,sizeof x)
#define write(x) printf("%d\n",x)
#define INF 0x3f3f3f3f
#define ll long long
#define mod  998244353
#define pii pair<int,int>
const int N = 1e5+ 5;
const int M =  2e5+5;
int head[N],tot;
struct Edge
{int next;int to;
}edge[M];
inline void add(int from,int to){edge[++tot].next = head[from];edge[tot].to = to;head[from] = tot;
}
int dfn[N],low[N],num;
bool bridge[N];
void tarjan(int x,int in_edge)
{dfn[x] = low[x] = ++num;for(int i = head[x];~i;i = edge[i].next){int y = edge[i].to;if(!dfn[y]){tarjan(y,i);low[x] = min(low[x],low[y]);if(low[y] > dfn[x]){bridge[i] = bridge[i ^ 1] = true;}}else if(i != (in_edge ^ 1)) low[x] = min(low[x],dfn[y]);}
}
bool w[2020][2020];
inline void init()
{mmt(head,-1);tot = 1;mmt(dfn,0);mmt(low,0);num = 0;mmt(bridge,0);mmt(w,0);
}
int main()
{//freopen("input.txt","r",stdin);int n;int u,v;int m;tot = 1;while(scanf("%d",&n) == 1){init();for(int i = 1;i <= n;++i){scanf("%d",&u);char o,p;cin>>o>>m>>p;for(int j = 1; j<= m;++j){read(v);if(w[u][v] == 0){w[u][v] = w[v][u] = 1;add(u,v);add(v,u);}// add(v,u);}}vector<pii> Q;for(int i = 1;i <= n;++i) if(!dfn[i]) tarjan(i,0);for(int i = 2;i <  tot;i += 2){if(bridge[i]){int a=  edge[i^1].to,b = edge[i].to;if(a > b) swap(a,b);Q.push_back({a,b});}}printf("%d critical links\n",Q.size());int Size = Q.size();for(int i = 0; i < Size;++i){printf("%d - %d\n",Q[i].first,Q[i].second);}cout<<endl;}
}