HDU Problem 3466 Proud Merchants【01背包】

Proud Merchants

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 5404    Accepted Submission(s): 2273

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 1010 15 105 10 53 105 10 53 5 62 7 3

 

Sample Output

511

 

Author

iSea @ WHU

 

Source

2010 ACM-ICPC Multi-University Training Contest（3）——Host by WHU

 

Recommend

zhouzeyong   |   We have carefully selected several similar problems for you:  3460 3463 3468 3467 3465 

 

比其他的01背包问题，这道题多了排序，而且排序按照data[i].q - data[i].p 升序排，比较难想到。

假设我们要买a (a.p, a.q) 和 b(b.p, b.q)两个物品。假设两者单独或者说没有限制条件q是都可以买到的，如果先买a,那么就需要 a.p + b.q 元，如果先买b, 就需要 b.p + a.q 元。因为在最开始剩下的钱数最多，所以我们希望开始的时候所需要的金额最大。所以a.p + b.q > a.q + b.p   ===》 a.q - a.p < b.q - b.p；

#include <cmath>#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>using namespace std;typedef __int64 Int;typedef long long LL;const double ESP = 1e-5;const double Pi = acos(-1.0);const int MAXN = 10000 + 10;const int INF = 0x3f3f3f3f;struct node{    int p, q, v;} data[MAXN];bool cmp(node a, node b) {    return a.q - a.p < b.q - b.p;}int dp[MAXN];int main() {    int n, m;    while (scanf("%d%d", &n, &m) != EOF) {        for (int i = 0; i < n; i++) {            scanf("%d%d%d", &data[i].p, &data[i].q, &data[i].v);        }        sort(data, data + n, cmp);        memset(dp, 0, sizeof(dp));        for (int i = 0; i < n; i++) {            for (int j = m; j >= data[i].p; j--) {                if (data[i].q > j) continue;                dp[j] = max(dp[j], dp[j - data[i].p] + data[i].v);            }        }        printf("%d\n", dp[m]);    }    return 0;}