PAT-A 1013 题解-优快云博客

本文链接：https://blog.youkuaiyun.com/ThinkingLion/article/details/44779461

源地址：http://www.patest.cn/contests/pat-a-practise/1013

题目大意就是给你n个城市以及m对城市之间两两有路径存在，然后k个询问，每个询问输入一个城市号，问如果去掉与这座城市相连的所有的路，问你至少修几条路，才能使得剩下的城市全部连通。。

其实很简单的，，去掉某个城市后，对剩余的图dfs一遍，或者用并查集记录，目的就是为了查询图中有几个独立的模块。

dfs:

#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<time.h>
#include<queue>
#include<stack>
#include<iterator>
#include<math.h>
#include<stdlib.h>
#include<limits.h>
#include<set>
#include<map>
//#define ONLINE_JUDGE
#define eps 1e-3
#define INF 0x7fffffff
#define FOR(i,a) for((i)=0;i<(a);(i)++)
#define MEM(a) (memset((a),0,sizeof(a)))
#define sfs(a) scanf("%s",a)
#define sf(a) scanf("%d",&a)
#define sfI(a) scanf("%I64d",&a)
#define pf(a) printf("%d\n",a)
#define pfI(a) printf("%I64d\n",a)
#define pfs(a) printf("%s\n",a)
#define sfd(a,b) scanf("%d%d",&a,&b)
#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)
#define for1(i,a,b) for(int i=(a);i<b;i++)
#define for2(i,a,b) for(int i=(a);i<=b;i++)
#define for3(i,a,b)for(int i=(b);i>=a;i--)
#define MEM1(a) memset(a,0,sizeof(a))
#define MEM2(a) memset(a,-1,sizeof(a))
const double PI=acos(-1.0);
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
using namespace std;
#define ll long long
int n,m;
#define Mod 1000000007
#define N 510
#define M 1000100
const int size = 10010;
const int mod = 9901;
int mp[1010][1010];
int vis[1010];
void dfs(int u){
	for(int i=0;i<n;i++){
		if(!vis[i] && mp[u][i]){
			vis[i] = 1;
			dfs(i);
		}
	}
}
int main(){
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
//  freopen("out.txt","w",stdout);
#endif
    int k;
    while(scanf("%d%d%d",&n,&m,&k)!=EOF){
    	memset(mp,0,sizeof mp);
    	int u,v;
    	for(int i=0;i<m;i++){
    		scanf("%d%d",&u,&v);
    		u--;
    		v--;
    		mp[u][v] = 1;
    		mp[v][u] = 1;
    	}
    	for(int i=0;i<k;i++){
    		memset(vis,0,sizeof vis);
    		scanf("%d",&u);
    		u--;
    		vis[u] = 1;
    		int num=0;
    		for(int j=0;j<n;j++){
    			if(!vis[j]){
    				dfs(j);
    				num++;
    			}
    		}
    		printf("%d\n",num-1);
    	}
    }
return 0;
}

并查集：

#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<time.h>
#include<queue>
#include<stack>
#include<iterator>
#include<math.h>
#include<stdlib.h>
#include<limits.h>
#include<set>
#include<map>
//#define ONLINE_JUDGE
#define eps 1e-3
#define INF 0x7fffffff
#define FOR(i,a) for((i)=0;i<(a);(i)++)
#define MEM(a) (memset((a),0,sizeof(a)))
#define sfs(a) scanf("%s",a)
#define sf(a) scanf("%d",&a)
#define sfI(a) scanf("%I64d",&a)
#define pf(a) printf("%d\n",a)
#define pfI(a) printf("%I64d\n",a)
#define pfs(a) printf("%s\n",a)
#define sfd(a,b) scanf("%d%d",&a,&b)
#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)
#define for1(i,a,b) for(int i=(a);i<b;i++)
#define for2(i,a,b) for(int i=(a);i<=b;i++)
#define for3(i,a,b)for(int i=(b);i>=a;i--)
#define MEM1(a) memset(a,0,sizeof(a))
#define MEM2(a) memset(a,-1,sizeof(a))
const double PI=acos(-1.0);
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
using namespace std;
#define ll long long
int n,m;
#define Mod 1000000007
#define N 510
#define M 1000100
const int size = 10010;
const int mod = 9901;
vector<int> edge[1010];
int fa[1010];
void Init(){
	for(int i=0;i<=n;i++){
		fa[i] = i;
	}
}
int findfa(int x){
	return x==fa[x]?fa[x]:fa[x]=findfa(fa[x]);
}
void Union(int x,int y){
	int fx = findfa(x);
	int fy = findfa(y);
	if(fx != fy)
		fa[fy] = fx;
}
int main(){
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
//  freopen("out.txt","w",stdout);
#endif
    int k;
    while(scanf("%d%d%d",&n,&m,&k)!=EOF){
    	for(int i=0;i<=n;i++)
    		edge[i].clear();
    		int u,v;
    		for(int i=0;i<m;i++){
    			scanf("%d%d",&u,&v);
    			u--;
    			v--;
    			edge[u].push_back(v);
    			edge[v].push_back(u);
    		}
    		while(k--){
    			Init();
    			scanf("%d",&u);
    			u--;
    			for(int i=0;i<n;i++){
    				int size = (int)edge[i].size();
    				if(i == u) continue;
    				for(int j=0;j<size;j++){
    					int v = edge[i][j];
    					if(v == u) continue;
    					Union(i,v);
    				}
    			}
    			int num=0;
    			for(int i=0;i<n;i++){
    				if(i==u) continue;
    				if(fa[i] == i) num++;
    			}
    			printf("%d\n",num-1);
    		}
    }
return 0;
}

pat_1013