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 }