uva10985

题目链接请戳 这里

 

解题思路

先用Floyed算出最短路。

枚举任意点对，得到最短路上的所有点。去掉连接同一层次点的边就好了。

 

代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N 140
#define INF 1e9
using namespace std;

int d[N][N], a[N][N], s[N];
int n;

void floyd()
{
    for (int k = 0; k < n; k++)
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++) if (d[i][k] < INF && d[k][j] < INF)
                d[i][j] = min(d[i][k] + d[k][j], d[i][j]);
}

int main()
{
    int t, c = 0;
    scanf("%d", &t);
    while (t--) {
        int m;
        scanf("%d%d", &n, &m);
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++) {
                d[i][j] = INF;
                a[i][j] = 0;
                if (i == j) { d[i][j] = 0; a[i][j] = 1; }
            }
        for (int i = 0; i < m; i++) {
            int x, y;
            scanf("%d%d", &x, &y);
            d[x][y] = d[y][x] = 1; a[x][y] = a[y][x] = 1;
        }
        //floyd预处理
        floyd();
        int maxn = 0;
        //枚举任意两点
        for (int i = 0; i < n; i++)
            for (int j = i + 1; j < n; j++) if (i != j && d[i][j] != INF) {
                int tot = 0;    //最短路上点的个数
                int temp = 0;    //当前这两点上绷紧的绳子数目
                //找最短路上的点，注意最短可能有多条
                for (int k = 0; k < n; k++) if (d[i][k] + d[k][j] ==  d[i][j]) s[tot++] = k;
                //枚举最短路上任意两点，去掉连接同一层次点的边
                for (int x = 0; x < tot; x++)
                    for (int y = x + 1; y < tot; y++) {
                        if (a[s[x]][s[y]] == 1 && d[i][s[x]] != d[i][s[y]]) temp++;
                    }
                maxn = max(maxn, temp);
            }
        printf("Case #%d: ", ++c);
        printf("%d\n", maxn);
    }
    return 0;
}