PAT-Top1002. Business (35)

　　在一个项目的截止日期之前，如果工期有空闲则可能可以开展其他项目，提高效益。本题考查动态规划。数组dp[i][t]表示在截止时间为t时，前i个项目工作安排能够产生的最大收益，而前i个项目的截止时间都不大于t。

//#include "stdafx.h"
#include <iostream>
#include <algorithm>
#include <memory.h>

using namespace std;

struct node { // project struct
    int p, l, d; // p is the profit, l is the lasting days of the project, d is the deadline
}pro[51];

int cmp(node a, node b) { // sort rule
    return a.d < b.d;
}

int main() {
    int n;
    scanf("%d", &n);

    int i, maxd = 0;
    for (i = 1; i <= n; i++) {
        scanf("%d%d%d", &pro[i].p, &pro[i].l, &pro[i].d);

        if (pro[i].d > maxd) { // get the max deadline
            maxd = pro[i].d;
        }
    }

    sort(pro + 1, pro + n + 1, cmp);

    int** dp = new int*[n + 1];
    for (i = 0; i <= n; i++) {
        dp[i] = new int[maxd + 1];
        memset(dp[i], 0, sizeof(dp[i])); // initialization : set value zero
    }

    //printf("%d
", dp[0][0]);
    int j, t;
    for (i = 1; i <= n; i++) {
        for (j = 1; j <= maxd; j++) {
            t = min(j, pro[i].d) - pro[i].l; 
            // get the max profit
            if (t >= 0) { // if can plus current project to compare the profit
                dp[i][j] = max(pro[i].p + dp[i - 1][t], dp[i - 1][j]);
            } else { // otherwise
                dp[i][j] = dp[i - 1][j];
            }
        }
    }

    printf("%d
", dp[n][maxd]);

    for (i = 0; i <= n; i++) {
        delete[] dp[i];
    }
    delete[] dp;

    system("pause");
    return 0;
}

参考资料

PAT. 1002. Business (35)