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[];

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

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

     int i, maxd = ;
     for (i = ; 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 + , pro + n + , cmp);

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

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

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

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

     system("pause");
     return ;
 }

参考资料

PAT. 1002. Business (35)