UVA11134-Fabled Rooks（贪心）

来源：互联网发布：定时继续 python 编辑：程序博客网时间：2024/06/05 15:47

题目链接

http://acm.hust.edu.cn/vjudge/problem/34086

题意

给定一个n x n的棋盘，在棋盘上摆放n个皇后，使其不相互攻击，并且对于每个皇后，给定一个矩阵区域，该皇后只能在这个矩形区域内，输出一个合法方案/impossible

思路

对于每个皇后，其摆放的x，y坐标是独立的，于是就将题意转化成了两个一维的问题：对于一个皇后，给定区间[l, r]，只能在该区间上选择一点。最终使1-n中的所有点均能被选择。
于是题目转化成了一个经典的贪心问题，类似于活动安排问题：
首先将区间按照右端点从早到晚排序，右端点相同的，按左端点从早到晚排序。然后用位置去覆盖每个点

代码

#include <iostream>#include <cstring>#include <stack>#include <vector>#include <set>#include <map>#include <cmath>#include <queue>#include <sstream>#include <iomanip>#include <fstream>#include <cstdio>#include <cstdlib>#include <climits>#include <deque>#include <bitset>#include <algorithm>using namespace std;#define PI acos(-1.0)#define LL long long#define PII pair<int, int>#define PLL pair<LL, LL>#define mp make_pair#define IN freopen("in.txt", "r", stdin)#define OUT freopen("out.txt", "wb", stdout)#define scan(x) scanf("%d", &x)#define scan2(x, y) scanf("%d%d", &x, &y)#define scan3(x, y, z) scanf("%d%d%d", &x, &y, &z)#define sqr(x) (x) * (x)#define pr(x) cout << #x << " = " << x << endl#define lc o << 1#define rc o << 1 | 1#define pl() cout << endl#define CLR(a, x) memset(a, x, sizeof(a))#define FILL(a, n, x) for (int i = 0; i < n; i++) a[i] = xconst int maxn = 5005;int dx[maxn], dy[maxn], n;struct node {    int l, r, id;    node(int a, int b, int c) : l(a), r(b), id(c) {    }    node() {    }} L[maxn], R[maxn];bool cmp(node x, node y) {    if (x.r != y.r) return x.r < y.r;    return x.l < y.l;}bool solve(node a[], int sta) {    int vis[maxn], pos = 1;    CLR(vis, 0);    for (int i = 0; i < n; i++) {        node &no = a[i];        for (int j = no.l; j <= no.r; j++) {            if (vis[j]) {                if (j == no.r) return false;                continue;            }               vis[j] = 1;            if (sta == 1) dx[no.id] = j;            if (sta == 2) dy[no.id] = j;            break;        }    }    return true;}int main() {    int x, y, z, w;    while (scan(n) && n) {        for (int i = 0; i < n; i++) {            scanf("%d%d%d%d", &x, &y, &z, &w);            L[i] = node(x, z, i);            R[i] = node(y, w, i);        }        sort(L, L + n, cmp);        sort(R, R + n, cmp);        bool f1 = solve(L, 1);        bool f2 = solve(R, 2);        if (!f1 || !f2)             puts("IMPOSSIBLE");        else             for (int i = 0; i < n; i++) cout << dx[i] << ' ' << dy[i] << endl;    }    return 0;}

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