题意 在n*n的棋盘上的n个指定区间上各放1个‘车’ 使他们相互不攻击 输出一种可能的方法
行和列可以分开看 就变成了n个区间上选n个点的贪心问题 看行列是否都有解就行 基础的贪心问题 对每个点选择包含它的最优未使用空间
#include <bits/stdc++.h> using namespace std; const int N = 5005; int xl[N], yl[N], xr[N], yr[N], x[N], y[N], n; bool solve(int a[], int l[], int r[]) { int cur, mr; //mr为包含k的区间最小右界,cur为放k的最优区间 memset(a, -1, sizeof(int)*n); for(int k = 1; k <= n; ++k) { cur = -1, mr = N; for(int i = 0; i < n; ++i) if(a[i] < 0 && l[i] <= k && r[i] < mr) mr = r[cur = i]; if(cur < 0 || k > mr) return 0; a[cur] = k; } return 1; } int main() { while(~scanf("%d", &n), n) { for(int i = 0; i < n; ++i) scanf("%d%d%d%d", &xl[i], &yl[i], &xr[i], &yr[i]); if(solve(x, xl, xr) && solve(y, yl, yr)) for(int i = 0; i < n; ++i) printf("%d %d\n", x[i], y[i]); else puts("IMPOSSIBLE"); } return 0; }
Fabled Rooks
We would like to place n rooks,
1 ≤ n ≤ 5000, on a n×n board
subject to the following restrictions
- The i-th rook can only be placed within the rectangle given by its left-upper corner (xli, yli) and its right-lower corner (xri, yri), where 1 ≤ i ≤ n,
1 ≤ xli ≤ xri ≤ n, 1 ≤ yli ≤ yri ≤ 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. n lines follow giving the rectangles where the rooks
can be placed as described above. The i-th line among them givesxli, yli, xri, and yri. 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 outputn lines 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. Output IMPOSSIBLE if there is no such placing of the rooks.
Sample input
8 1 1 2 2 5 7 8 8 2 2 5 5 2 2 5 5 6 3 8 6 6 3 8 5 6 3 8 8 3 6 7 8 8 1 1 2 2 5 7 8 8 2 2 5 5 2 2 5 5 6 3 8 6 6 3 8 5 6 3 8 8 3 6 7 8 0
Output for sample input
1 1 5 8 2 4 4 2 7 3 8 5 6 6 3 7 1 1 5 8 2 4 4 2 7 3 8 5 6 6 3 7