并查集 判环

小希的迷宫

Description

上次Gardon的迷宫城堡小希玩了很久（见Problem B），现在她也想设计一个迷宫让Gardon来走。但是她设计迷宫的思路不一样，首先她认为所有的通道都应该是双向连通的，就是说如果有一个通道连通了房间A和B，那么既可以通过它从房间A走到房间B，也可以通过它从房间B走到房间A，为了提高难度，小希希望任意两个房间有且仅有一条路径可以相通（除非走了回头路）。小希现在把她的设计图给你，让你帮忙判断她的设计图是否符合她的设计思路。比如下面的例子，前两个是符合条件的，但是最后一个却有两种方法从5到达8。

Input

输入包含多组数据，每组数据是一个以0 0结尾的整数对列表，表示了一条通道连接的两个房间的编号。房间的编号至少为1，且不超过100000。每两组数据之间有一个空行。
整个文件以两个-1结尾。

Output

对于输入的每一组数据，输出仅包括一行。如果该迷宫符合小希的思路，那么输出”Yes”，否则输出”No”。

Sample Input

6 8 5 3 5 2 6 4
5 6 0 0

8 1 7 3 6 2 8 9 7 5
7 4 7 8 7 6 0 0

3 8 6 8 6 4
5 3 5 6 5 2 0 0

-1 -1

Sample Output

Yes
Yes
No

Hint

题意

题解：

就是判断是不是一棵树 前提不能有环的存在

AC代码

#include<cstdio>
#include<cstring>
#include<stack>
#include <set>
#include <queue>
#include <vector>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
const int mod = 1e9+7;
int vis[100005];
int par[100005];
int x,y;
int flag,edge;
void init(){
    for (int i = 1;i <= 100005; ++i){
        par[i] = i;
        edge = 0;
        flag = 0;
        vis[i] = 0;
    }
}
int fd(int x){
    if (x == par[x]) return x;
    return par[x]=fd(par[x]);
}
void ut(int x,int y){
    if (x==y)
    {
        flag = 1; return ;
    }
    int tx = fd(x);
    int ty = fd(y);
    if (tx!=ty){
       par[tx] = ty;
       edge++;
    }
    else flag = 1;
}
int main()
{
    int mx;
    while (scanf("%d%d",&x,&y),x!=-1||y!=-1){
            if (x==0&&y==0)
            {
                printf("Yes\n"); continue;
            }
           init();
           vis[x] = 1;
           vis[y] = 1;
           ut(x,y);
            mx = -1;
            mx = max(mx,x);
            mx = max(mx,y);
            while (scanf("%d%d",&x,&y),x!=0||y!=0){
                ut(x,y);
                vis[x] = 1;
                vis[y] = 1;
                mx = max(mx,x);
                mx = max(mx,y);
            }
            int v = 0;
            for (int i = 1;i <= mx; ++i){
                    if (vis[i]) v++;
            }
            if (!flag&&edge+1==v) printf("Yes\n");
            else printf("No\n");
    }
    return 0;
}

Ice_cream’s world I

Description

ice_cream’s world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are set up, and wall between watchtowers be build, in order to partition the ice_cream’s world. But how many ACMers at most can be awarded by the queen is a big problem. One wall-surrounded land must be given to only one ACMer and no walls are crossed, if you can help the queen solve this problem, you will be get a land.

Input

In the case, first two integers N, M (N<=1000, M<=10000) is represent the number of watchtower and the number of wall. The watchtower numbered from 0 to N-1. Next following M lines, every line contain two integers A, B mean between A and B has a wall(A and B are distinct). Terminate by end of file.

Output

Output the maximum number of ACMers who will be awarded.
One answer one line.

Sample Input

8 10
0 1
1 2
1 3
2 4
3 4
0 5
5 6
6 7
3 6
4 7

Sample Output

3

Hint

题意

题解：

判断分成区块的个数 其实是判断环

AC代码

#include<cstdio>
#include<cstring>
#include<stack>
#include <set>
#include <queue>
#include <vector>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
const int mod = 1e9+7;
int par[1005];
int x,y;
int flag,cnt;
void init(){
    for (int i = 0;i < 1005; ++i){
        par[i] = i;
    }
}
int fd(int x){
    if (x == par[x]) return x;
    return par[x]=fd(par[x]);
}
void ut(int x,int y){
    int tx = fd(x);
    int ty = fd(y);
    if (tx!=ty){
       par[tx] = ty;
    }
    else cnt++;
}
int main()
{
    int n,m;
    while (scanf("%d%d",&n,&m)!=EOF){
        init();
        cnt = 0;
        int x,y;
        for (int i = 0;i < m; ++i){
            scanf("%d%d",&x,&y);
            ut(x,y);
        }
        printf("%d\n",cnt);
    }

    return 0;
}