题目:
http://acm.hdu.edu.cn/showproblem.php?pid=5222
题意:
给一个图,图中有若干有向边,若干无向边,有重边,无自环,问图中是否存在环
思路:
分类考虑
先判断无向边中是否存在环,
若不存在,则仅由 点 与 无向边 组成的图,为若干独立的连通子图,且均为生成树;
再考虑有向边,
若存在一有向边弧头和弧尾属于同一连通子图,则必然该子图内存在存在环
若不存在,则将每个连通子图抽象为一个节点,判断有向图中是否存在环
判断无向图中的环可以用并查集,有向图中的环用拓扑排序即可
代码:
#include<iostream>
#include<vector>
#include<stdio.h>
#include<queue>
#include<string.h>
#include<map>
using namespace std;
const int MAXSIZE = 1000005;
map<int,int> node_arr;
vector<int> adjlist[MAXSIZE];
int s_set[MAXSIZE];
int s_rank[MAXSIZE];
int num[MAXSIZE];
int s_find(int x){
int k = x, t;
while (k != s_set[k])
k = s_set[k];
while (x != k){
t = s_set[x];
s_set[x] = k;
x = t;
}
return x;
}
void s_union(int x, int y){
x = s_find(x);
y = s_find(y);
if (x == y) return;
if (s_rank[x] >= s_rank[y]){
s_set[y] = x;
if (s_rank[x] == s_rank[y]) s_rank[x]++;
num[x] += num[y];
}
else{
s_set[x] = y;
num[y] += num[x];
}
return;
}
int main(){
int total;
cin>>total;
while (total--){
int n,m1,m2;
cin>>n>>m1>>m2;
for (int i=1;i<=n;++i){
adjlist[i].clear();
}
for (int i = 0; i <= n; ++i){
s_set[i] = i;
s_rank[i] = num[i] = 1;
}
bool isLoop = false;
for (int i=1;i<=m1;++i){
int a,b;
scanf("%d %d",&a,&b);
if (s_find(a) ==s_find(b)){
isLoop=true;
}
else s_union(a,b);
}
node_arr.clear();
for (int i=1;i<=m2;++i){
int a,b;
scanf("%d %d",&a,&b);
if (s_find(a) == s_find(b)) isLoop=true;
adjlist[s_find(a)].push_back(s_find(b));
node_arr[s_find(a)];
node_arr[s_find(b)]++;
}
//in
if (isLoop) cout<<"YES"<<endl;
else {
// for (int i=1;i<=n;++i)
// cout<<"i:"<<i<<" "<<s_find(i)<<endl;
queue<int> q;
for (map<int,int>::iterator it=node_arr.begin();it!=node_arr.end(); ++it){
//cout<<it->first<<" "<<it->second<<endl;
if (it->second==0){
q.push(it->first);
it->second=-1;
}
}
while (!q.empty()){
int t=q.front();
// cout<<"q:"<<t<<endl;
q.pop();
for (int i=0;i<adjlist[t].size();++i){
if ((--node_arr[adjlist[t][i]])==0){
q.push(adjlist[t][i]);
node_arr[adjlist[t][i]]=-1;
}
}
}
for (map<int,int>::iterator it=node_arr.begin();it!=node_arr.end();++it){
// cout<<it->first<<" "<<it->second<<endl;
if (it->second!=-1){
isLoop=true;
break;
}
}
if (isLoop) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
}
return 0;
}