pku 1018 Communication System dp

#include <iostream>using namespace std;int v[105][1001];int Max = 1000;int main(){int N, n, m;int wide, price;scanf("%d", &N);while(N--){scanf("%d", &n);memset(v, 0x7f, sizeof(v));//dpfor(int i = 1; i <= Max; ++i)v[0][i] = 0;for(int i = 1; i <= n; ++i){scanf("%d", &m);for(int j = 0; j < m; ++j){scanf("%d%d", &wide, &price);for(int k = 1; k <= wide; ++k)if(v[i-1][k] + price < v[i][k])v[i][k] = v[i-1][k] + price;for(int k = wide+1; k <= Max; ++k)if(v[i-1][k] + price < v[i][wide])v[i][wide] = v[i-1][k] + price;}}//cal sumdouble ret = 0.0;for(int i = 1; i <= Max; ++i)if((1.0*i/v[n][i]) > ret)ret = 1.0*i/v[n][i];printf("%.3lf/n", ret);}return 0;}//dp问题和经典的解答方法://v[i][j]为前i件物品,width为j时的最小price//i:1->n,不断更新width为j时的最小price