hiho一下 第九十週 Swimming Plans

時間限制:10000ms

單點時限:1000ms

內存限制:256MB

描述

Steven loves swimming! The swimming pool he goes to every afternoon consists of N parallel lanes which are number from 0 to N-1. Steven plans to swim for R rounds. Each round he chooses a lane and swim to the other side, spending exactly L units of time.

Since the weather is getting hotter, the swimming pool is very crowded. To prevent from accidents, the swimming pool manager asks everyone to submit their plans before swimming. A plan contains specific informations for a round, including t - the time he starts, l - the time he spends, n - the lane he chooses, d - the direction he swims(0 for swimming from west to east and 1 for swimming from east to west). Everyone starts at some integeral time and swims at a constant speed.

The plans should not conflict with each other: two swimmers are not allowed to meet in the middle of a lane (meeting at the sides is allowed).

Steven arrives at time T and gets Q plans of the other swimmers. So here's the question, what's the earliest time that Steven can finish his swimming plans?

Note that Steven starts from the west side and he moves from one side to the other only by swimming.

輸入

Input contains multiple testcases.

The first line is an integer TASKS, representing the number of testcases.

For each testcase, the first line contains five integers T, L, R, N and Q which are described above.

The following Q lines describe the plans, each line with four integers, t, l, n, d, which are described above.

For all testcases:

0 <= T, t <= 10⁴

1 <= L, l <= 10⁴

1 <= R <= 10⁴

0 <= Q <= 10⁴

1 <= N <= 10³

輸出

The earliest time that Steven can finish his swimming plans.

題目分析

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

typedef struct Event{
	int t;
	int x;
	int n;
}Event;

Event event[2][20000];
int count[2];

void AddEvent(int t, int x, int n, int d){
	event[d][count[d]].t = t;
	event[d][count[d]].x = x;
	event[d][count[d]].n = n;
	count[d]++;
}

int comp(const void* a, const void* b){
	Event* c = (Event*)a;
	Event* d = (Event*)b;
	
	if(c->t != d->t)
		return c->t - d->t;
	else
		return c->x - d->x;	
}

int main(){
	int TASK, T, L, R, N, Q;
	int t, l, n, d;
	int i, D, find, point[2], lane[2][1000];
	
	scanf("%d", &TASK);
	while(TASK--){
		count[0] = 0;
		count[1] = 0;
		scanf("%d%d%d%d%d", &T, &L, &R, &N, &Q);
		for(i=0; i<Q; i++){
			scanf("%d%d%d%d", &t, &l, &n, &d);
			if(l < L){
				AddEvent(t-L+l, 1, n, d);
				AddEvent(t, -1, n, d);
			}else if(l == L){
				AddEvent(t-1, 1, n, d);
				AddEvent(t+1, -1, n, d);
			}else{
				AddEvent(t, 1, n, d);
				AddEvent(t+l-L, -1, n, d);
			}
			
			AddEvent(t-L, 1, n, 1-d);
			AddEvent(t+l, -1, n, 1-d);
		}
		
		qsort(event[0], Q*2, sizeof(Event), comp);
		qsort(event[1], Q*2, sizeof(Event), comp);
		
		D = 0;
		point[0] = 0;
		point[1] = 0;
		memset(lane, 0, sizeof(lane));
		while(R){
			find = 0;
			while(point[D] < 2*Q){
				if(event[D][point[D]].t < T){
					lane[D][event[D][point[D]].n] += event[D][point[D]].x;
					point[D]++;
				}else{
					for(i=0; i<N; i++){
						if(lane[D][i] == 0){
							find = 1;
							T += L;
							break;
						}
					}
					if(find)
						break;
					T = event[D][point[D]].t;
					lane[D][event[D][point[D]].n] += event[D][point[D]].x;
					point[D]++;	
				}	
			}
			if(!find)
				T += L;
			D = 1 - D;
			R--;	
		}
		
		printf("%d\n", T);
	}
	
	return 0;
} 

這道題A得好辛苦，memset那句，用sizeof(int)*2*N就會出錯

今天先不管了，明天再想

有人能給解釋下嗎？