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]

 1 /*
 2  State:Accepted
 3  Time:2013.3.2
 4 */
 5 
 6 #include <iostream>
 7 #include <cstring>
 8 #include <string>
 9 #include <fstream>
10 #include <cstdlib>
11 #include <cstdio>
12 #include <algorithm>
13 #include <cmath>
14 int dp[200] , sum[200] , p[200] ,test , n ;
15 
16 int min(int x , int y){
17       return x < y? x : y;
18 }
19 
20 void solve(){
21      scanf("%d",&n);
22      memset(sum , 0 ,sizeof(sum));
23      for (int i = 1; i <= n; ++i)
24        {
25            scanf("%d%d",&sum[i] , &p[i]);
26            sum[i] += sum[i - 1];
27        }
28      for (int i = 1; i <= n; ++i){
29         dp[i] = 1000000000;
30         for (int j = 0; j < i; ++j)
31              dp[i] = min(dp[i] , dp[j] + (sum[i] - sum[j] + 10) * p[i]);
32      }
33      printf("%d\n",dp[n]);
34      
35 }
36 
37 int main(){
38      freopen("poj1260.in","r",stdin);
39      freopen("poj1260.out","w",stdout);
40      scanf("%d",&test);
41      while ( test -- ){
42            solve();
43      }
44 }

 

posted on 2013-03-23 20:35  yzcstc  阅读(137)  评论(0)  编辑  收藏  举报