九度OJ 1325：Battle Over Cities（城市间的战争） （并查集）

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：376

解决：132

题目描述：

It is vitally important to have all the cities connected by highways in a war. If a city is occupied by the enemy, all the highways from/toward that city are closed. We must know immediately if we need to repair any other highways to keep the rest of the cities
connected. Given the map of cities which have all the remaining highways marked, you are supposed to tell the number of highways need to be repaired, quickly.

For example, if we have 3 cities and 2 highways connecting city₁-city₂ and city₁-city₃.
Then if city₁ is occupied by the enemy, we must have 1 highway repaired, that is the highway city₂-city₃.

输入：

Each input file contains one test case. Each case starts with a line containing 3 numbers N (<1000), M and K, which are the total number of cities, the number of remaining highways, and the number of cities to be checked, respectively. Then M lines follow,
each describes a highway by 2 integers, which are the numbers of the cities the highway connects. The cities are numbered from 1 to N. Finally there is a line containing K numbers, which represent the cities we concern.

输出：

For each of the K cities, output in a line the number of highways need to be repaired if that city is lost.

样例输入：

3 2 3
1 2
1 3
1 2 3

样例输出：

1
0
0

思路：

题意是求，若第i个城市被占领了，需要另外修几条路。

我的思路是对每种情况求并查集，求集合个数num，num-1就是需要修的路。

代码：

#include <stdio.h>

#define N 1000
#define M 1000000

typedef struct node {
    int x;
    int y;
} ROAD;

int n;
int pre[N+1];
int count[N+1];
int num;

void init()
{
    for (int i=1; i<=n; i++)
    {
        pre[i] = i;
        count[i] = 1;
    }
    num = n;
}

int find(int i)
{
    while (i != pre[i])
        i = pre[i];
    return i;
}

int combine(int i, int j)
{
    int a = find(i);
    int b = find(j);
    if (a != b)
    {
        if (count[a] > count[b])
        {
            pre[b] = a;
            count[a] += count[b];
            count[b] = 0;
        }
        else
        {
            pre[a] = b;
            count[b] += count[a];
            count[a] = 0;
        }
        num --;
        return 1;
    }
    else
        return 0;
}

int main(void)
{
    int m, k, i, j;
    ROAD r[M];
    int concern[N+1];

    while (scanf("%d%d%d", &n, &m, &k) != EOF)
    {
        for(i=0; i<m; i++)
            scanf("%d%d", &r[i].x, &r[i].y);
        for(j=0; j<k; j++)
        {
            init();
            scanf("%d", &concern[j]);
            for(i=0; i<m; i++)
            {
                if (r[i].x != concern[j] && r[i].y != concern[j])
                    combine(r[i].x, r[i].y);
                if (num == 2)
                    break;
            }
            printf("%d\n", num-2);
        }
    }

    return 0;
}
/**************************************************************
    Problem: 1325
    User: liangrx06
    Language: C
    Result: Accepted
    Time:190 ms
    Memory:8664 kb
****************************************************************/