POJ1337 Balance（完全揹包～～）

A Lazy Worker

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 1131		Accepted: 384

Description

There is a worker who may lack the motivation to perform at his peak level of efficiency because he is lazy. He wants to minimize the amount of work he does (he is Lazy, but he is subject to a constraint that he must be busy when there is work that he can do.) 

We consider a set of jobs 1, 2,..., n having processing times t1, t2,...,tn respectively. Job i arrives at time ai and has its deadline at time di. We assume that ti, ai, and di have nonnegative integral values. The jobs have hard deadlines, meaning that each job i can only be executed during its allowed interval Ii=[ai, di]. The jobs are executed by the worker, and the worker executes only one job at a time. Once a job is begun, it must be completed without interruptions. When a job is completed, another job must begin immediately, if one exists to be executed. Otherwise, the worker is idle and begins executing a job as soon as one arrives. You should note that for each job i, the length of Ii, di - ai, is greater than or equal to ti, but less than 2*ti. 

Write a program that finds the minimized total amount of time executed by the worker. 

Input

The input consists of T test cases. The number of test cases (T ) is given in the first line of the input file. The number of jobs (0<=n<=100) is given in the first line of each test case, and the following n lines have each job's processing time(1<=ti<=20),arrival time(0<=ai<=250), and deadline time (1<=di<=250) as three integers.

Output

Print exactly one line for each test case. The output should contain the total amount of time spent working by the worker.

Sample Input

3
3
15 0 25
50 0 90
45 15 70
3
15 5 20
15 25 40
15 45 60
5
3 3 6
3 6 10
3 14 19
6 7 16
4 4 11

Sample Output

50
45
15

題意：有一個天平，若干砝碼，要用上所有砝碼，使天平處於平衡狀態，問有幾種不同情況。

當放上去的砝碼不同時，天平會有不同的平衡狀態，設爲j，當j<0時表示左邊重；當j>0時表示右邊重；當j=0時則天平處於平衡。

爲了使數組下標不出現負數情況，要用到狀態轉移（見代碼），所以，當j=7500時爲平衡狀態。

/*完全揹包*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int c,g,hook[25],weight[25];
int dp[25][15050];
//dp[i][j]表示放上第i個後的平衡係數爲j的possobilities。
//爲了使數組下標不出現負數情況，狀態轉移。[-20*15*25 ~ 20*15*25]->[0 ~ 2*20*15*25]即[0 ~ 15000]
//dp[i][7500]是平衡狀態

int main()
{
    //freopen("in.txt","r",stdin);
    memset(dp,0,sizeof(dp));
    scanf("%d%d",&c,&g);
    for(int i = 0; i < c; i++)
        scanf("%d",&hook[i]);
    for(int i= 0; i < g; i++)
        scanf("%d",&weight[i]);
    dp[0][7500]=1;
    for(int i = 0; i < g; i++)
    {
        for(int j = 0; j <= 15000; j++)
        {
            if(dp[i][j])
                for(int k = 0; k < c; k++)
                    dp[i + 1][j + weight[i] * hook[k]] += dp[i][j];
        }
    }
    printf("%d\n",dp[g][7500]);
}