百度之星2017资格赛1003 度度熊与邪恶大魔王

思路：
真是菜的不行。
实现：
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 
 6 const int INF = 0x3f3f3f3f;
 7 int dp[11][1001], A[100005], B[100005], K[1005], P[1005], n, m; 
 8 
 9 int main()
10 {
11     while (scanf("%d %d", &n, &m) != EOF)
12     {
13         for (int i = 0; i < n; i++) scanf("%d %d", &A[i], &B[i]);
14         for (int i = 0; i < m; i++) scanf("%d %d", &K[i], &P[i]);
15         memset(dp, 0x3f, sizeof dp);
16         for (int j = 0; j <= 10; j++)
17         {
18             dp[j][0] = 0;
19             for (int i = 0; i < m; i++)
20             {
21                 if (P[i] > j)
22                 {
23                     for (int k = 1; k <= 1000; k++)
24                     {
25                         dp[j][k] = min(dp[j][k], K[i] * ((k + P[i] - j - 1) / (P[i] - j)));
26                     }
27                 }
28             }
29         }
30         
31         for (int j = 0; j <= 10; j++)
32         {
33             for (int i = 1; i < m; i++)
34             {
35                 for (int k = 1; k <= 1000; k++)
36                 {
37                     if (P[i] > j && k - (P[i] - j) >= 0)
38                     {
39                         dp[j][k] = min(dp[j][k], dp[j][k - (P[i] - j)] + K[i]);
40                     }
41                 }
42             }
43         }
44         
45         long long sum = 0;
46         bool flg = true;
47         for (int i = 0; i < n; i++)
48         {
49             if (dp[B[i]][A[i]] >= INF)
50             {
51                 flg = false; break;
52             }
53             sum += (long long)dp[B[i]][A[i]];
54         }
55         if (!flg) puts("-1");
56         else printf("%I64d
", sum);
57     }
58     return 0;
59 }