pat 1013. Battle Over Cities (25)

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本文介绍了一种用于确定战时因城市被占领而导致高速公路网断裂时所需的修复路径数量的算法。通过使用并查集数据结构，该算法能够快速计算出在特定城市丢失的情况下，为保持剩余城市间的连接性所需修复的高速公路数量。

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.

Input

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.

Output

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

Sample Input
3 2 3
1 2
1 3
1 2 3
Sample Output
1
0
0

这题我用的是并查集，每次都构建一个并查集，父节点是自己的节点个数便是连通分量的个数cnt。去掉的节点是个连通分量，其余的有cnt-1个连通分量需要cnt-2条路。当然 cnt可能是1或0；这题M很大，我开十万的爆了，开百万才行。

#include<cstdio>
#include<cstring>
using namespace std;

int p[10000];
int findp(int x)
{
    if (x!=p[x])
        p[x]=findp(p[x]);
    return p[x];
}
void merge(int x,int y)
{
    int px=findp(x);
    int py=findp(y);
    if (px==py) return;
    p[px]=py;
}
int main()
{
    int n,i,j,k,m;
    int x[1000000],y[1000000];
    scanf("%d%d%d",&n,&m,&k);
    for (i=0;i<m;i++)
        scanf("%d%d",&x[i],&y[i]);
    for (i=0;i<k;i++)
    {
        int now;
        scanf("%d",&now);
        for (j=1;j<=n;j++)
            p[j]=j;
        for (j=0;j<m;j++)
        {
            if (x[j]!=now&&y[j]!=now)
                merge(x[j],y[j]);
        }
        int cnt=0;
        for (j=1;j<=n;j++)
        {
            if (p[j]==j)
                cnt++;
        }
        cnt-=2;
        if (cnt<0) cnt=0;
        printf("%d\n",cnt);
    }
    return 0;
}