hiho一下第九十一周 Events Arrangement-优快云博客

本文链接：https://blog.youkuaiyun.com/ZzebraA/article/details/51068865

本文讨论了如何在给定时间内通过合理安排活动来最大化游行的兴奋度，并确保至少有一分钟的空闲时间。包括输入参数解析、排序策略、动态规划算法的应用以及时间约束下的最优解求取。

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时间限制: 20000ms

单点时限: 1000ms

内存限制: 256MB

描述

There are n events in the parade. Each one can be performed for at most k times.

The parade will last for m minutes. The major will give a talk at the end, so you should make at least one minute idle.

Each events have three attributions, a_i, b_i, c_i. If the parade have p minutes remaining when the i-th event begins, the excitement of the parade will increase by a_i-(m-p)*b_i and the event will last for c_i minutes.

You should arrange the events to maximize excitement and guarantee at least one minute at the end will be free.

输入

Three integers n, m ,k on the first line.

The following n lines, there are three integers a_i, b_i, c_i, denoting the attributions of the events.

0 <= a_i, b_i, c_i <= 10000

For 20% of the data, n<=5, m<=10, k<=5；

For 60% of the data, n<=200, m<=500, k<=50；

For 100% of the data, n<=1000, m<=5000, k<=1000

输出

One integer on one line, denoting the answer.

题目分析

TLE代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h> 

int dp[1000][500];

struct Event{
	int a, b, c;
	double d;
}event[1001];

int calc(int s, int e, int i){
	int j = (e-s)/event[i].c;
	return event[i].a*j - event[i].b*(j*s + event[i].c*(j-1)*j/2);
}

int cmp(const void* a, const void* b){
	return ((struct Event*)a)->d < ((struct Event*)b)->d ? 1:-1;
}

int main(){
	int n, m, k, i, tempvalue, max, v, j;
	
	scanf("%d%d%d", &n, &m, &k);
	for(i = 1; i <= n; i++){
		scanf("%d%d%d", &event[i].a, &event[i].b, &event[i].c);
		event[i].d = (double)event[i].b/(double)event[i].c;
	}
	
	qsort(&event[1], n, sizeof(struct Event), cmp);
	
	memset(dp, 0, sizeof(dp)); 
	for(i = 1; i <= n; i++){
		for(v = 0; v < m; v++){
			for(j = 0; j <= k && j*event[i].c <= v; j++){
				tempvalue = calc(v-event[i].c*j, v, i);
				if(dp[i][v] < dp[i-1][v-event[i].c*j] + tempvalue)
					dp[i][v] = dp[i-1][v-event[i].c*j] + tempvalue;	
			}
		}
	}
	for(max = dp[n][0], i = 1; i < m; i++){
		if(max < dp[n][i])
			max = dp[n][i];
	}
	printf("%d\n", max);
	return 0;
}

AC代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h> 

int dp[1001][5000], q[10000][500];

struct Event{
	int a, b, c;
	double d;
}event[1001];

int calc(int s, int e, int i){
	int j = (e-s)/event[i].c;
	return event[i].a*j - event[i].b*(j*s + event[i].c*(j-1)*j/2);
}

int cmp(const void* a, const void* b){
	return ((struct Event*)a)->d < ((struct Event*)b)->d ? 1:-1;
}

int main(){
	int n, m, k, i, tempvalue, max, v, j, tail[10000], head[10000], r;
	int tempvalue1, tempvalue2, value1, value2;
	
	scanf("%d%d%d", &n, &m, &k);
	for(i = 1; i <= n; i++){
		scanf("%d%d%d", &event[i].a, &event[i].b, &event[i].c);
		event[i].d = (double)event[i].b/(double)event[i].c;
	}
	
	qsort(&event[1], n, sizeof(struct Event), cmp);
	
	memset(dp, 0, sizeof(dp)); 
	for(i = 1; i <= n; i++){
		memset(tail, 0, sizeof(tail));
		for(j = 0; j < event[i].c; j++){
			head[j] = 1;
		}
		for(v = 0; v < m; v++){
			r = v%event[i].c;
			tail[r]++;
			q[r][tail[r]] = v;
			
			while(tail[r] > head[r]){
				tempvalue1 = calc(q[r][tail[r]], v, i);
				value1 = dp[i-1][q[r][tail[r]]] + tempvalue1;
				
				tempvalue2 = calc(q[r][tail[r]-1], v, i);
				value2 = dp[i-1][q[r][tail[r]-1]] + tempvalue2;
				
				if(value1 >= value2){
					q[r][tail[r]-1] = q[r][tail[r]];
					tail[r]--;
				}else{
					break;
				}
			}
			while((v - q[r][head[r]])/event[i].c > k){
				head[r]++;
			}
			tempvalue = calc(q[r][head[r]], v, i);
			dp[i][v] = dp[i-1][q[r][head[r]]] + tempvalue;
		}
	}
	
	for(max = dp[n][0], i = 1; i < m; i++){
		if(max < dp[n][i])
			max = dp[n][i];
	}
	printf("%d\n", max);
	
	return 0;
}

为何要排序？

因为枚举时每加入一个新的event，总是把它放在最后