UVA 11134 Fabled Rooks（传说中的车）（贪心）

题意：在n*n的棋盘上放n个车，使得任意两个车不相互攻击，且第i个车在一个给定的矩形Ri之内，不相互攻击是指不同行不同列，无解输出IMPOSSIBLE，否则分别输出第1,2，……，n个车的坐标。

分析：行和列是无关的，因此把原题分解成两个一维问题。在区间[1,n]内选择n个不同的整数，使得第i个整数在闭区间[n1i, n2i]内。按r优先排序。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 5000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
bool vis[MAXN];
int ans[MAXN][3];
int n;
struct Node{
    int l, r, id;
    Node(){}
    bool operator < (const Node& a)const{
        return r < a.r;
    }
}num1[MAXN], num2[MAXN];
bool judge(Node *num, int x){
    memset(vis, false, sizeof vis);
    sort(num + 1, num + n + 1);
    for(int i = 1; i <= n; ++i){
        bool ok = false;
        for(int j = num[i].l; j <= num[i].r; ++j){
            if(!vis[j]){
                vis[j] = true;
                ans[num[i].id][x] = j;
                ok = true;
                break;
            }
        }
        if(!ok) return false;
    }
    return true;
}
int main(){
    while(scanf("%d", &n) == 1){
        if(!n) return 0;
        memset(ans, 0, sizeof ans);
        for(int i = 1; i <= n; ++i){
            scanf("%d%d%d%d", &num1[i].l, &num2[i].l, &num1[i].r, &num2[i].r);
            num1[i].id = num2[i].id = i;
        }
        if(!judge(num1, 0)){
            printf("IMPOSSIBLE\n");
            continue;
        }
        if(!judge(num2, 1)){
            printf("IMPOSSIBLE\n");
            continue;
        }
        for(int i = 1; i <= n; ++i){
            printf("%d %d\n", ans[i][0], ans[i][1]);
        }
    }
    return 0;
}