【HDU1507】【最大匹配输出路径match数组】
来源:互联网 发布:农村淘宝订单佣金 编辑:程序博客网 时间:2024/06/15 16:26
Uncle Tom's Inherited Land*
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2505 Accepted Submission(s): 1034
Special Judge
Problem Description
Your old uncle Tom inherited a piece of land from his great-great-uncle. Originally, the property had been in the shape of a rectangle. A long time ago, however, his great-great-uncle decided to divide the land into a grid of small squares. He turned some of the squares into ponds, for he loved to hunt ducks and wanted to attract them to his property. (You cannot be sure, for you have not been to the place, but he may have made so many ponds that the land may now consist of several disconnected islands.)
Your uncle Tom wants to sell the inherited land, but local rules now regulate property sales. Your uncle has been informed that, at his great-great-uncle's request, a law has been passed which establishes that property can only be sold in rectangular lots the size of two squares of your uncle's property. Furthermore, ponds are not salable property.
Your uncle asked your help to determine the largest number of properties he could sell (the remaining squares will become recreational parks).
Your uncle Tom wants to sell the inherited land, but local rules now regulate property sales. Your uncle has been informed that, at his great-great-uncle's request, a law has been passed which establishes that property can only be sold in rectangular lots the size of two squares of your uncle's property. Furthermore, ponds are not salable property.
Your uncle asked your help to determine the largest number of properties he could sell (the remaining squares will become recreational parks).
Input
Input will include several test cases. The first line of a test case contains two integers N and M, representing, respectively, the number of rows and columns of the land (1 <= N, M <= 100). The second line will contain an integer K indicating the number of squares that have been turned into ponds ( (N x M) - K <= 50). Each of the next K lines contains two integers X and Y describing the position of a square which was turned into a pond (1 <= X <= N and 1 <= Y <= M). The end of input is indicated by N = M = 0.
Output
For each test case in the input your program should first output one line, containing an integer p representing the maximum number of properties which can be sold. The next p lines specify each pair of squares which can be sold simultaneity. If there are more than one solution, anyone is acceptable. there is a blank line after each test case. See sample below for clarification of the output format.
Sample Input
4 461 11 42 24 14 24 44 344 23 22 23 10 0
Sample Output
4(1,2)--(1,3)(2,1)--(3,1)(2,3)--(3,3)(2,4)--(3,4)3(1,1)--(2,1)(1,2)--(1,3)(2,3)--(3,3)
Source
South America 2002 - Practice
Recommend
LL
#include <iostream>#include <cstring>#include <cmath>#include <queue>#include <stack>#include <list>#include <map>#include <set>#include <string>#include <cstdlib>#include <cstdio>#include <algorithm>using namespace std;int n,m,k;const int N = 510;int g[N][N];int a[N][N];int b[N*N];int used[N];int match[N];int uN,vN;bool dfs(int u) { for(int i=1;i<=vN;i++) { if(!used[i] && g[u][i]) { used[i] = true; if(match[i] == -1 || dfs(match[i])) { match[i] = u; return true; } } } return false; } int hungary() { int ret = 0; for(int i=1;i<=uN;i++) { memset(used,0,sizeof(used)); if(dfs(i)) ret++; } return ret; }int main(){ freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); while(scanf("%d%d",&n,&m) ){if(n == 0 && m == 0) break;scanf("%d",&k);int index = 1;memset(a,0,sizeof(a));for(int i=0;i<k;i++){int x,y;scanf("%d%d",&x,&y);x --;y --;a[x][y] = -1;}for(int i=0;i<n;i++){for(int j=0;j<m;j++){if(a[i][j] == -1) continue;b[index] = i * m + j;a[i][j] = index ++;}}memset(g,0,sizeof(g));memset(match,-1,sizeof(match));//cout << "ok" << endl;for(int i=0;i<n;i++){for(int j=0;j<m;j++){if(a[i][j] != -1 && (i+j) % 2 == 1){if(i + 1 < n && a[i+1][j] != -1){g[a[i][j]][a[i+1][j]] = 1;}if(i - 1 >= 0 && a[i-1][j] != -1){g[a[i][j]][a[i-1][j]] = 1;}if(j + 1 < m && a[i][j+1] != -1){g[a[i][j]][a[i][j+1]] = 1;}if(j - 1 >= 0 && a[i][j-1] != -1){g[a[i][j]][a[i][j-1]] = 1;}}}}//cout << "ok" << endl;uN = vN = index - 1;int ret = hungary();printf("%d\n",ret);for(int i=1;i<=vN;i++){if(match[i] != -1){int x1 = b[i] / m; int y1 = b[i] % m;int x2 = b[match[i]] / m;int y2 = b[match[i]] % m;printf("(%d,%d)--(%d,%d)\n",x1+1,y1+1,x2+1,y2+1);}}printf("\n");} return 0;}
0 0
- 【HDU1507】【最大匹配输出路径match数组】
- HDU1507(最大二分匹配)
- hdu1507--二分图最大匹配
- hdu1507 Uncle Tom's Inherited Land*--最大匹配
- hdoj 2819 Swap (最大匹配+输出路径)
- hdu1507(二分匹配)
- HDU 2819 Swap (最大二分匹配+输出路径)
- hdu 2819 Swap(二分图最大匹配,输出路径)
- hdu2819 Swap 二分图最大匹配 输出路径
- HDU1507-二分图行列匹配
- HDU3081 Marriage Match II 【最大匹配】
- hdu1507
- Hdu1507
- 【bzoj1264】【基因匹配Match】【dp+树状数组】
- ZOJ1516&&HDU1507(二分图匹配)
- HDOJ 题目2189 Swap(二分图最大匹配,输出路径)
- HDU I'm Telling the Truth (二分图最大匹配+字典序最大路径输出(好题))
- HDU--3081--Marriage Match II--最大匹配,匈牙利算法
- scala macro annotation 使用 例子
- hdoj 1042 N! 【大数阶乘】
- Linux下推荐应用程序列表【2008-07-31】
- CCF 201312-4 有趣的数 (数位DP)
- 一张图看懂 why 代理和数据源 使用 weak 修饰
- 【HDU1507】【最大匹配输出路径match数组】
- 文章标题
- Jquery ui Interactions方法的使用实例
- Kinect v2.0原理介绍之七:彩色帧获取
- 在VS2012下创建单元测试项目
- 基本数据类型的转换
- 机器学习实战
- MANACHER 最长回文序列
- PCA and Whitening Exercise