HDU1507-二分圖行列匹配

2019-11-08 19:34:22

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Uncle Tom's Inherited Land*

Time Limit: 2000/1000 MS (java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3423 Accepted Submission(s): 1438Special JudgePRoblem DescriptionYour 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).

InputInput 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. OutputFor 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 Output4(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) 題目大意：
                   給你一個n*m的矩陣，其中有些格子不能填東西，然后問你在空白處最多能填多少個1*2的方塊
題目思路：
                 首先我們想到1*2的方塊就是相鄰兩個格子的匹配，所以我們可以考慮將相鄰格子建邊，做最大匹配，最后的答案就是我們要求的，然后我們考慮如何構造二分圖，應為這是個二維坐標，所以我們為了方便可以把二維坐標轉換成一維坐標，然后枚舉所有格子，分成奇偶，行列相加為偶數的和行列相加為奇數的連邊，因為兩個相鄰格子的奇偶不同，然后進行最大匹配，集合的大小就是n*m，最后最大匹配的答案就是我們要求的，對于輸出邊，我們可以從link數組當中記錄的連接邊的情況來輸出，我們枚舉i，
如果link[i]不等于-1，這時候我們就知道i和link[i]之間有一條邊
AC代碼：
#include<cstring>#include<cstdio>#include<vector>using std::vector;const int maxn = 2e4+100;bool vis[maxn],mp[205][205];int link[maxn];int n,m,k;vector<int>edge[maxn];bool dfs(int u){    for(int i=0;i<edge[u].size();i++){        int v = edge[u][i];        if(!vis[v]){            vis[v]=true;            if(link[v]==-1||dfs(link[v])){                link[v]=u;                return true;            }        }    }    return false;}void graph(){    for(int i=1;i<=n*m;i++)edge[i].clear();    for(int i=1;i<=n;i++){        for(int j=1;j<=m;j++){            int x = (i-1)*m+j;            if(!mp[i][j]&&(i+j)%2){                if(j+1<=m&&!mp[i][j+1]){         //枚舉偶數點的四個方向                    edge[x].push_back(x+1);                }                if(i+1<=n&&!mp[i+1][j]){                    edge[x].push_back(x+m);                }                if(j-1>=1&&!mp[i][j-1]){                    edge[x].push_back(x-1);                }                if(i-1>=1&&!mp[i-1][j]){                    edge[x].push_back(x-m);                }            }        }    }}int main(){    while(scanf("%d%d",&n,&m),n+m){        scanf("%d",&k);        memset(link,-1,sizeof(link));        memset(mp,false,sizeof(mp));        for(int j=1;j<=k;j++)        {            int x,y;scanf("%d%d",&x,&y);            mp[x][y]=true;        }        graph();        int ans = 0;        for(int i=1;i<=n*m;i++){            memset(vis,false,sizeof(vis));            if(dfs(i))ans++;        }        printf("%d/n",ans);        if(ans>0){            for(int i=1;i<=n*m;i++){    //枚舉連接的邊                int j = link[i];                if(link[i]!=-1){                    int x = i,y = j;                    printf("(%d,%d)--(%d,%d)/n",(x-1)/m+1,(x-1)%m+1,(y-1)/m+1,(y-1)%m+1);                }            }        }        printf("/n");    }    return 0;}