HDU 2464

    N<=8，顯然枚舉，時間複雜度O(N!*N*N)。

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <iostream>
#include <sstream>

using namespace std;

const int inf = 1000000000;
bool ga[8][8], gb[8][8];
int idx[8];

int main() {
    int n, ea, eb, ia, ib, da, db;
    int ans, i, x, y, m, tmp, cas = 1;
    while (scanf("%d %d %d", &n, &ea, &eb), n || ea || eb) {
        memset(ga, 0, sizeof(ga));
        memset(gb, 0, sizeof(gb));
        scanf("%d %d %d %d", &ia, &ib, &da, &db);
        for (i = 0; i < ea; i++) {
            scanf("%d %d", &x, &y);
            ga[x][y] = ga[y][x] = 1;
        }
        for (i = 0; i < eb; i++) {
            scanf("%d %d", &x, &y);
            gb[x][y] = gb[y][x] = 1;
        }
        m = 1;
        for (i = 0; i < n; i++) {
            idx[i] = i;
            m *= i + 1;
        }
        ans = inf;
        for (i = 0; i < m; i++) {
            tmp = 0;
            for (x = 0; x < n; x++) {
                for (y = x + 1; y < n; y++) {
                    if (ga[x][y] != gb[idx[x]][idx[y]]) {
                        if (ga[x][y]) {
                            tmp += min(da, ib);
                        } else {
                            tmp += min(ia, db);
                        }
                    }
                }
            }
            if (tmp < ans) ans = tmp;
            next_permutation(idx, idx + n);
        }
        printf("Case #%d: %d\n", cas++, ans);
    }
    return 0;
}