poj1260

思路：dp

     f[i][j]表示前i个总共买了j个总共要付的最小费用。。。

     f[i][j] = min(f[i -1][k] + (j-k+10)*value[i])；  k<=sum[i-1]

/*
 State:Accepted
 Time:2013.3.2
*/

#include <iostream>
#include <cstring>
#include <string>
#include <fstream>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <cmath>
int dp[200] , sum[200] , p[200] ,test , n ;

int min(int x , int y){
      return x < y? x : y;
}

void solve(){
     scanf("%d",&n);
     memset(sum , 0 ,sizeof(sum));
     for (int i = 1; i <= n; ++i)
       {
           scanf("%d%d",&sum[i] , &p[i]);
           sum[i] += sum[i - 1];
       }
     for (int i = 1; i <= n; ++i){
        dp[i] = 1000000000;
        for (int j = 0; j < i; ++j)
             dp[i] = min(dp[i] , dp[j] + (sum[i] - sum[j] + 10) * p[i]);
     }
     printf("%d\n",dp[n]);
     
}

int main(){
     freopen("poj1260.in","r",stdin);
     freopen("poj1260.out","w",stdout);
     scanf("%d",&test);
     while ( test -- ){
           solve();
     }
}