HDU 3433 (DP + 二分) A Task Process

 1 //#define LOCAL
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5 using namespace std;
 6 
 7 int dp[55][205], a[205], b[205];
 8 int _, n, x, y, kase = 0;
 9 
10 bool check(int t)
11 {
12     memset(dp, -1, sizeof(dp));
13     dp[0][0] = 0;
14 
15     for(int i = 1; i <= n; ++i)
16     {
17         if(dp[i][x] >= y)    return true;    
18         for(int j = 0; j <= x; ++j)
19         {
20             if(dp[i - 1][j] != -1)
21             {
22                 int temp = min(t/a[i], x-j);    //枚举第i个人可以做的A任务的个数
23                 for(int k = 0; k <= temp; ++k)
24                 {
25                     int t1 = (t - a[i] * k) / b[i];    //计算第i个人做k件A任务，所能做的B任务的个数
26                     dp[i][j + k] = max(dp[i][j + k], dp[i - 1][j] + t1);    //选最优解
27                 }
28             }
29         }
30     }
31 
32     if(dp[n][x] >= y)    return true;
33     return false;
34 }
35 
36 int main(void)
37 {
38     #ifdef LOCAL
39         freopen("3433in.txt", "r", stdin);
40     #endif
41 
42     scanf("%d", &_);
43     while(_--)
44     {
45         scanf("%d%d%d", &n, &x, &y);
46         for(int i = 1; i <= n; ++i)
47             scanf("%d%d", &a[i], &b[i]);
48 
49         int l = 0, r = a[1] * x + b[1] * y;
50         int ans = r;
51         while(l <= r)
52         {
53             int mid = (l + r) >> 1;
54             if(check(mid))
55             {
56                 ans = mid;
57                 r = mid - 1;
58             }
59             else    l = mid + 1;
60         }
61 
62         printf("Case %d: %d\n", ++kase, ans);
63     }
64 
65     return 0;
66 }