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?
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.
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
題意：首先給出物品數量和手中資金
人後每樣物品給出價格，需要購買時手中至少需要多少資金，還有物品本身的價值
要求求出最大資金
思路：01背包，要注意的是要先按q-p排序
原本我是按其價值排序的，WA一次
[cpp]
#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，保證差額最小為最優
{
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--)//剩余的錢大於q才能買
{
dp[j] = max(dp[j],dp[j-a[i].p]+a[i].v);//這裡的j-a[i].p決定了之前的排序方法
}
}
printf("%d\n",dp[m]);
}
return 0;
}
#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，保證差額最小為最優
{
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--)//剩余的錢大於q才能買
{
dp[j] = max(dp[j],dp[j-a[i].p]+a[i].v);//這裡的j-a[i].p決定了之前的排序方法
}
}
printf("%d\n",dp[m]);
}
return 0;
}
