hdu5445 || 2015长春网络赛1009题 多重背包问题

http://acm.hdu.edu.cn/showproblem.php?pid=5445

Problem Description

Few days before a game of orienteering, Bell came to a mathematician to solve a big problem. Bell is preparing the dessert for the game. There are several different types of desserts such as small cookies, little grasshoppers and tiny mantises. Every type of dessert may provide different amounts of energy, and they all take up different size of space.

Other than obtaining the desserts, Bell also needs to consider moving them to the game arena. Different trucks may carry different amounts of desserts in size and of course they have different costs. However, you may split a single dessert into several parts and put them on different trucks, then assemble the parts at the game arena. Note that a dessert does not provide any energy if some part of it is missing.

Bell wants to know how much would it cost at least to provide desserts of a total energy of  p  (most of the desserts are not bought with money, so we assume obtaining the desserts costs no money, only the cost of transportation should be considered). Unfortunately the mathematician is having trouble with her stomach, so this problem is left to you.

 

Input

The first line of input contains a integer  T(T≤10)  representing the number of test cases.

For each test case there are three integers  n,m,p  on the first line  (1≤n≤200,1≤m≤200,0≤p≤50000) , representing the number of different desserts, the number of different trucks and the least energy required respectively.

The  i−th  of the  n  following lines contains three integers  ti,ui,vi(1≤ti≤100,1≤ui≤100,1≤vi≤100)  indicating that the  i−th  dessert can provide  ti energy, takes up space of size  ui  and that Bell can prepare at most  vi  of them.

On each of the next  m  lines, there are also three integers  xj,yj,zj(1≤xj≤100,1≤yj≤100,1≤zj≤100)  indicating that the  j−th  truck can carry at most size of  xj  , hiring each one costs  yj  and that Bell can hire at most  zj  of them.

 

Output

For every test case output the minimum cost to provide the dessert of enough energy in the game arena if it is possible and its cost is no more than  50000 . Otherwise, output  TAT  on the line instead.

 

Sample Input

   
   
   
   

    
    
    
    4
1 1 7
14 2 1
1 2 2
1 1 10
10 10 1
5 7 2
5 3 34
1 4 1
9 4 2
5 3 3
1 3 3
5 3 2
3 4 5
6 7 5
5 3 8
1 1 1
1 2 1
1 1 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    4
14
12
TAT
   
   
   
   
   
   
   
   


   
   
   
   

/**
hdu5445 || 2015长春网络赛1009题   多重背包问题
题目大意：给一场运动会提供食物，每种食物提供ti能量，占用vi空间，最多可提供ui个，把食物运到指定地点，每种车可以运送ai体积的
          食物，消耗bi的金钱，总共有ci个这种车，问给运动会提供至少p的能量，最少需要花多少运费
          特别说明：每个食物可以拆开来运
解题思路：利用多重背包的二进制方案，先找出提供不少于p的能量需要多少空间V（背包必须全部装满），然后求运送能力大于V的最小背包
*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=100500;
int n,m,p;
int c[maxn],w[maxn],dp[maxn];

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&p);
        int nn=0,t,u,v;
        for(int i=0; i<n; i++)
        {
            scanf("%d%d%d",&t,&u,&v);
            int k=0;
            while(v-(1<<(k+1))+1>0)k++;
            for(int j=0; j<k; j++)
            {
                ++nn;
                c[nn]=t*(1<<j);
                w[nn]=u*(1<<j);
            }
            ++nn;
            c[nn]=t*(v-(1<<k)+1);
            w[nn]=u*(v-(1<<k)+1);
        }
        memset(dp,0x3f3f3f3f,sizeof(dp));
        dp[0]=0;
        for(int i=1; i<=nn; i++)
        {
            for(int j=p+100; j>=c[i]; j--)//单个食物提供能量最大100，因此我们找p~p+100之间的完全装满的最小体积
            {
                dp[j]=min(dp[j],dp[j-c[i]]+w[i]);
            }
        }
        int V=0x3f3f3f3f;
        for(int i=p; i<=p+100; i++)V=min(dp[i],V);
        nn=0;
        for(int i=0; i<m; i++)
        {
            scanf("%d%d%d",&u,&t,&v);
            int k=0;
            while(v-(1<<(k+1))+1>0)k++;
            for(int j=0; j<k; j++)
            {
                ++nn;
                c[nn]=t*(1<<j);
                w[nn]=u*(1<<j);
            }
            ++nn;
            c[nn]=t*(v-(1<<k)+1);
            w[nn]=u*(v-(1<<k)+1);
        }
        memset(dp,0,sizeof(dp));
        for(int i=1; i<=nn; i++)
        {
            for(int j=50000; j>=c[i]; j--)///最大体积不超过50000
            {
                dp[j]=max(dp[j],dp[j-c[i]]+w[i]);
            }
        }
        int ans = 0x3f3f3f3f;
        for(int i=50000; i>=0; i--)
        {
            if (dp[i] >= V)///找运送体积过V的最小背包即为答案
            {
                ans = min(ans, i);
            }
        }
        if (ans == 0x3f3f3f3f)
            puts("TAT");
        else
            printf("%d\n", ans);
    }
    return 0;
}