Treasure Map Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 3209 Appoint description:  System Crawler  (2015-04-09) Description Your boss once had got many copies of a treasure map. Unfortunately…

Treasure Map Time Limit: 2 Seconds      Memory Limit: 32768 KB Your boss once had got many copies of a treasure map. Unfortunately, all the copies are now broken to many rectangular pieces, and what make it worse, he has lost some of the pieces. Luck…

直接精确覆盖 开始逐行添加超时了,换成了单点添加 #include <iostream> #include <cstring> #include <cstdio> #include <cmath> #include <vector> using namespace std; #define FOR(i,A,s) for(int i = A[s]; i != s; i = A[i]) #define exp 1e-8 ; int n, m, k,…

题目链接 精确覆盖的模板题, 把每一个格子当成一列就可以. S忘记初始化TLE N次, 哭晕在厕所...... #include<bits/stdc++.h> using namespace std; #define pb(x) push_back(x) #define ll long long #define mk(x, y) make_pair(x, y) #define lson l, m, rt<<1 #define mem(a) memset(a, 0, sizeof(a…

题意:一个 n x m 的矩形(1 <= n, m <= 30),现给出这个矩形中 p 个(1 <= p <= 500)子矩形的左下角与右下角坐标,问最少用多少个子矩形能够恰好组成这个 n x m 的大矩形. 题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3372 -->>这是精确覆盖问题,而DLX正是解决精确覆盖问题的有利武器.. 模型转换:将原矩形变成一行,作为 DLX 中的列,表示要…

Description Your boss once had got many copies of a treasure map. Unfortunately, all the copies are now broken to many rectangular pieces, and what make it worse, he has lost some of the pieces. Luckily, it is possible to figure out the position of e…

Treasure Map Time Limit: 2 Seconds      Memory Limit: 32768 KB Your boss once had got many copies of a treasure map. Unfortunately, all the copies are now broken to many rectangular pieces, and what make it worse, he has lost some of the pieces. Luck…

题意 :  给你一个大小为 n * m 的矩形 , 坐标是( 0 , 0 ) ~ ( n , m )  .然后给你 p 个小矩形 . 坐标是( x1 , y1 ) ~ ( x2 , y2 ) , 你选择最小的几个矩形 , 使得这些矩形能够覆盖整个矩形 . 而且互相不会重叠 .( n , m <= 30 ) 思路 : Dancing Links 的精确覆盖问题 . 我们将 n * m 的矩形分成 n * m 个小正方形 ,那么我们仅仅要保证每一个小正方形被覆盖且仅仅被覆盖一次就可以 . 那么列表示…

精确覆盖 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <vector> using namespace std; const int maxn=920; const int maxnode=920*550; const int maxr=550; int ans; struct DLX { int n , sz; //…

用最少的矩阵覆盖n*m的地图.注意矩阵不能互相覆盖. 这里显然是一个精确覆盖,但因为矩阵拼接过程中,有公共的边,这里须要的技巧就是把矩阵的左边和以下截去一个单位. #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #…