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));

//dp
for(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 sum
double ret = 0.0;
for(int i = 1; i <= Max; ++i)
if((1.0*i/v
[i]) > ret)
ret = 1.0*i/v
[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