POJ - 1018 Communication System (暴力)

题目大意：要买n个零件，每个零件可以由m个厂家提供，每个零件都有相应的b值和p值。（每个零件只能买一个）现在要求你求出最大的min(b) /sum(p)
min(b)表示的是每个零件的最小b值
sum(p)表示的是每个零件的p值的和

解题思路：直接暴力

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define maxn 110
struct device{
    int b, p;
}D[maxn][maxn];
int num[maxn];
int n;

int cmp(const device a, const device b) {
    return a.p < b.p;
}

void solve() {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%d", &num[i]);
        for (int j = 0; j < num[i]; j++) 
            scanf("%d%d", &D[i][j].b, &D[i][j].p);
        sort(&D[i][0], &D[i][0] + num[i], cmp);
    }

    double ans = 0.0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < num[i]; j++) {
            int min_b = D[i][j].b;
            int sum_p = D[i][j].p;
            int k;

            for (k = 0; k < n; k++) {
                if (k == i)
                    continue;
                int tmp = 0;
                while (tmp < num[k] && D[k][tmp].b < min_b)
                    tmp++;
                if (tmp >= num[k])
                    break;
                sum_p += D[k][tmp].p;
            }
            if(k >= n && (double)min_b / (double)sum_p > ans)
                ans = (double)min_b  / (double)sum_p;
        }
    printf("%0.3f\n", ans);
}

int main() {
    int test;
    scanf("%d", &test);
    while(test--) {
        solve();
    }
    return 0;
}