//time时间内 dp[i][j]前i个人中完成j个A任务后最多能完成多少B任务
// dp[i][j] = max(dp[i][j],(time - k*a[i])/b[i]+dp[i-1][j-k]);
//由于time越大，dp[i][j]越大,故可以用二分找到最小的time的值
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 210;
const int inf = 0x7fffffff;
int dp[maxn][maxn];//time时间内 dp[i][j]前i个人中完成j个A任务后最多能完成多少B任务
int N,X,Y;
int a[maxn],b[maxn];
int DP(int time)
{
memset(dp,-1,sizeof(dp));
for(int j = 0;j*a[1] <= time && j<=X;j++)
dp[1][j] = (time-a[1]*j)/b[1];
for(int i = 2;i <= N;i++)
for(int j = 0;j <= X;j++)
for(int k = 0;k*a[i] <= time && k <= j;k++)
{
if(dp[i-1][j-k] == -1)
continue;
dp[i][j] = max(dp[i][j],(time - k*a[i])/b[i]+dp[i-1][j-k]);
}
if(dp[N][X] >= Y)
return 1;
else
return 0;
}
int bit(int left,int right)
{
while(left <= right)
{
int middle = (left+right)/2;
if(DP(middle))
right = middle - 1;
if(!DP(middle))
left = middle + 1;
}
return left;
}
int main()
{
//freopen("in.txt","r",stdin);
int T;
int cas = 0;
scanf("%d", &T);
while(T--)
{
scanf("%d%d%d",&N,&X,&Y);
int mi = inf ;
for(int i = 1;i <= N;i++)
{
scanf("%d%d",&a[i],&b[i]);
mi = min(a[i]*X+b[i]*Y,mi);
}
printf("Case %d: ",++cas);
printf("%d\n",bit(1,mi));
}
return 0;
}
hdu3433 A Task Process dp+二分
原文：http://blog.csdn.net/cq_pf/article/details/44628721
