HDU 3466 - Proud Merchants (01背包+sort)
2017-11-09 21:23
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Proud Merchants
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 7161 Accepted Submission(s): 2973
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
Sample Output
5
11
Author
iSea @ WHU
题意:
背包,加一个限制。
POINT:
q-p 从小到大排。死活半天想不出。
#include <stdio.h> #include <map> #include<iostream> #include <math.h> #include <algorithm> #include <string.h> #include <vector> #include <queue> #include <set> using namespace std; typedef long long LL; const int maxn = 600+55; int dp[6000]; struct node { int p,v,q; double val; }a[maxn]; bool cmd(node a ,node b) { return a.q-a.p<b.q-b.p; } int main() { int n,m; while(scanf("%d %d",&n,&m)!=EOF){ memset(dp,0,sizeof dp); for(int i=1;i<=n;i++){ scanf("%d %d %d",&a[i].p,&a[i].q,&a[i].v); a[i].val=1.0*a[i].v/a[i].p; } sort(a+1,a+1+n,cmd); for(int i=1;i<=n;i++){ for(int j=m;j>=a[i].q&&j>=a[i].p;j--){ dp[j]=max(dp[j],dp[j-a[i].p]+a[i].v); } } printf("%d\n",dp[m]); } return 0; }
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