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LOJ-10100(割点个数)

作者:互联网

题目链接:传送门

思路:

就是求割点的个数,直接Tarjan算法就行。

注意输入格式(判断比较水)。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<string>
#include<stack>
#include<algorithm>
using namespace std;
const int maxn = 550;
int num[maxn],low[maxn],vis[maxn],gedian[maxn],tim,cnt,root;
vector <int> vc[maxn];
int MIN(int x,int y)
{
    return x<y?x:y;
}
void Init()
{
    memset(num,0,sizeof(num));
    memset(low,0,sizeof(low));
    memset(vis,0,sizeof(vis));
    memset(gedian,0,sizeof(gedian));
    for(int i=0;i<maxn;i++) vc[i].clear();
    tim=0;cnt=0;
}
void Tarjan(int u,int pre)
{
    low[u]=num[u]=++tim;
    vis[u]=1;
    int v,i,tt=0;
    for(i=0;i<vc[u].size();i++){
        v=vc[u][i];
        if(!vis[v]){
            tt++;
            Tarjan(v,u);
            low[u]=MIN(low[u],low[v]);
            if((u==root&&tt>1)||(u!=root&&num[u]<=low[v])) gedian[u]=1;
        }
        else low[u]=MIN(low[u],num[v]);
    }
}
int main(void)
{
    int n,m,i,j,x,y;
    char ch;
    string str;
    while(~scanf("%d",&n)&&n){
        Init();
        int c1=0,c2;
        while(scanf("%d",&x)&&x){
            if(c1>=n) break;c1++;
            c2=0;
            while(1){
                c2++;
                scanf("%d",&y);
                vc[x].push_back(y);
                vc[y].push_back(x);
                if(c2>=n) break;
                ch=getchar();
                if(ch=='\n') break;
            }
        }
        
        for(i=1;i<=n;i++)
        if(vis[i]==0){
            root=i;
            Tarjan(i,-1);
        }
        for(i=1;i<=n;i++)
        if(gedian[i]==1) cnt++;
        printf("%d\n",cnt);
    }
    return 0;
} 
View Code

 

标签:vc,LOJ,割点,break,int,maxn,c2,include,10100
来源: https://www.cnblogs.com/2018zxy/p/10359296.html