Problem Description
Recently, iSea went to an ancient country. For such a long time, it was the most wealthy and powerful kingdom in the world. As a result, the people in this country are still very proud even if their nation hasn’t been so
wealthy any more.
The merchants were the most typical, each of them only sold exactly one item, the price was Pi, but they would refuse to make a trade with you if your money were less than Qi, and iSea evaluated every item a value Vi.
If he had M units of money, what’s the maximum value iSea could get?
The merchants were the most typical, each of them only sold exactly one item, the price was Pi, but they would refuse to make a trade with you if your money were less than Qi, and iSea evaluated every item a value Vi.
If he had M units of money, what’s the maximum value iSea could get?
Input
There are several test cases in the input.
Each test case begin with two integers N, M (1 ≤ N ≤ 500, 1 ≤ M ≤ 5000), indicating the items’ number and the initial money.
Then N lines follow, each line contains three numbers Pi, Qi and Vi (1 ≤ Pi ≤ Qi ≤ 100, 1 ≤ Vi ≤ 1000), their meaning is in the description.
The input terminates by end of file marker.
Each test case begin with two integers N, M (1 ≤ N ≤ 500, 1 ≤ M ≤ 5000), indicating the items’ number and the initial money.
Then N lines follow, each line contains three numbers Pi, Qi and Vi (1 ≤ Pi ≤ Qi ≤ 100, 1 ≤ Vi ≤ 1000), their meaning is in the description.
The input terminates by end of file marker.
Output
For each test case, output one integer, indicating maximum value iSea could get.
Sample Input
2 10 10 15 10 5 10 5 3 10 5 10 5 3 5 6 2 7 3
题意：首先给出物品数量和手中资金
人后每样物品给出价格，需要购买时手中至少需要多少资金，还有物品本身的价值
要求求出最大资金
思路：01背包，要注意的是要先按q-p排序
原本我是按其价值排序的，WA一次
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
struct node
{
int p,q,v;
} a[555];
int cmp(node x,node y)//按q-p排序，保证差额最小为最优
{
return x.q-x.p<y.q-y.p;
}
int main()
{
int n,m,i,j;
int dp[5555];
while(~scanf("%d%d",&n,&m))
{
for(i = 0; i<n; i++)
scanf("%d%d%d",&a[i].p,&a[i].q,&a[i].v);
memset(dp,0,sizeof(dp));
sort(a,a+n,cmp);
for(i = 0; i<n; i++)
{
for(j = m; j>=a[i].q; j--)//剩余的钱大于q才能买
{
dp[j] = max(dp[j],dp[j-a[i].p]+a[i].v);//这里的j-a[i].p决定了之前的排序方法
}
}
printf("%d\n",dp[m]);
}
return 0;
}
作者：libin56842 发表于2013-6-6 1:19:37 原文链接
阅读：41 评论：0 查看评论