Milking Time POJ - 3616 (DP最长上升子序列)

ゞ浴缸里的玫瑰 2022-02-25 11:48 174阅读 0赞

Bessie is such a hard-working cow. In fact, she is so focused on maximizing her productivity that she decides to schedule her next *N* (1 ≤ *N* ≤ 1,000,000) hours (conveniently labeled 0..*N*\-1) so that she produces as much milk as possible.

Farmer John has a list of *M* (1 ≤ *M* ≤ 1,000) possibly overlapping intervals in which he is available for milking. Each interval *i* has a starting hour (0 ≤ *starting\_houri* ≤ *N*), an ending hour (*starting\_houri* < *ending\_houri* ≤ *N*), and a corresponding efficiency (1 ≤ *efficiencyi* ≤ 1,000,000) which indicates how many gallons of milk that he can get out of Bessie in that interval. Farmer John starts and stops milking at the beginning of the starting hour and ending hour, respectively. When being milked, Bessie must be milked through an entire interval.

Even Bessie has her limitations, though. After being milked during any interval, she must rest *R* (1 ≤ *R* ≤ *N*) hours before she can start milking again. Given Farmer Johns list of intervals, determine the maximum amount of milk that Bessie can produce in the *N* hours.

Input

\* Line 1: Three space-separated integers: *N*, *M*, and *R*  
\* Lines 2..*M*\+1: Line *i*\+1 describes FJ's ith milking interval withthree space-separated integers: *starting\_houri* , *ending\_houri* , and *efficiencyi*

Output

\* Line 1: The maximum number of gallons of milk that Bessie can product in the *N* hours

Sample Input

12 4 2
    1 2 8
    10 12 19
    3 6 24
    7 10 31

Sample Output

解题思路:

先按照开始时间进行排序,然后求最大的能够连续的区间个数(最长上升子序列)

#include <iostream>
    #include <algorithm>
    #include <string.h>
    using namespace std;
    const int MAXM = 1010;  
    struct tinter {  //区间
    	int st, et, effi;  
    }tinters[MAXM];
    int N, M, R;  
    int mdp[MAXM]; //递推方程式  
    
    bool mcmp(tinter a, tinter b) {
    	if (a.st != b.st) {
    		return a.st < b.st;
    	}
    	else {
    		return a.et < b.et;
    	}
    }
    
    int main() {
    	cin >> N >> M >> R;  //R是休息时间,M是开始时间
    	for (int i = 0; i < M; ++i) {   //设定从1开始
    		cin >> tinters[i].st >> tinters[i].et >> tinters[i].effi;
    	}
    	memset(mdp, 0, sizeof(mdp));
    	sort(tinters, tinters + M, mcmp);  //先排个序
    	mdp[0] = tinters[0].effi;
    	for (int i = 1; i < M; ++i) {
    		
    		mdp[i] = tinters[i].effi;   //注意:终点也要单独取值!用于比较
    
    		for (int j = 0; j < i; ++j) {   //以i为终点时的状态
    			if (tinters[i].st >= tinters[j].et + R) {  
    				mdp[i] = max(mdp[i], mdp[j] + tinters[i].effi);
    			}
    		}
    	}
    	cout << *max_element(mdp, mdp + M) << endl;
    	
    	system("PAUSE");
    	return 0;
    }