UVA 11134 - Fabled Rooks(贪心+优先队列)

最新推荐文章于 2025-08-12 18:06:10 发布

最新推荐文章于 2025-08-12 18:06:10 发布 · 65 阅读

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解决在n×n棋盘上放置n辆车的问题，确保每辆车处于特定范围内且不互相攻击。采用贪心算法结合优先队列的方法进行求解。

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We would like to placenrooks, 1≤n≤5000, on an×nboard subject to the following restrictions

Thei-th rook can only be placed within the rectangle given by its left-upper corner (xl_i,yl_i) and its right-lower corner (xr_i,yr_i), where 1≤i≤n, 1≤xl_i≤xr_i≤n, 1≤yl_i≤yr_i≤n.
No two rooks can attack each other, that is no two rooks can occupy the same column or the same row.

The input consists of several test cases. The first line of each of them contains one integer number,n, the side of the board.nlines follow giving the rectangles where the rooks can be placed as described above. Thei-th line among them givesxl_i,yl_i,xr_i, andyr_i. The input file is terminated with the integer `0' on a line by itself.

Your task is to find such a placing of rooks that the above conditions are satisfied and then outputnlines each giving the position of a rook in order in which their rectangles appeared in the input. If there are multiple solutions, any one will do. OutputIMPOSSIBLEif there is no such placing of the rooks.

Sample input

Output for sample input

题意：给定n辆车可以放在棋盘的区间，求能否放满且所有车不受攻击。

思路：贪心+优先队列，把行列分开去贪心。。

代码：

#include <stdio.h>
#include <string.h>
#include <queue>
#include <algorithm>
using namespace std;

const int N = 5005;

int n;
struct Car {
    int xl, xr, yl, yr;
    int px, py, id;
} car[N];

bool cmp1(Car a, Car b) {
    if (a.yl != b.yl)
	return a.yl < b.yl;
    return a.yr < b.yr;
}

bool cmp2(Car a, Car b) {
    if (a.xl != b.xl)
	return a.xl < b.xl;
    return a.xr < b.xr;
}

bool cmp3(Car a, Car b) {
    return a.id < b.id;
}

void init() {
    for (int i = 0; i < n; i ++) {
	scanf("%d%d%d%d", &car[i].xl, &car[i].yl, &car[i].xr, &car[i].yr);
	car[i].id = i;
    }
}

bool solve() {
    sort(car, car + n, cmp1);
    for (int i = 1; i <= n; i ++) {
	if (car[i - 1].yl > i || car[i - 1].yr < i) return false;
	car[i - 1].py = i;
    }
    sort(car, car + n, cmp2);
    for (int i = 1; i <= n; i ++) {
	if (car[i - 1].xl > i || car[i - 1].xr < i) return false;
	car[i - 1].px = i;
    }
    sort(car, car + n, cmp3);
    return true;
}

int main() {
    while (~scanf("%d", &n) && n) {
	init();
	if (solve()) {
	    for (int i = 0; i < n; i ++)
		printf("%d %d\n", car[i].px, car[i].py);
	}
	else printf("IMPOSSIBLE\n");
    }
    return 0;
}