HDU 1507 Uncle Tom's Inherited Land*(二分图最大匹配:输出一组解)
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题意:N*M的矩形,向其中填充1*2的小块矩形,黑色的部分不能填充,白色部分需要填充且1*2矩形不能重叠。问最多可以填充多少块。输出任意一组填充解即可.
思路:首先同样将原图的所有可填充部分小方格编号,然后将这些合法小方格(可填充)按照他们的行号+列号是奇数还是偶数分为两个点集.(有一个重要的结论,矩阵或者棋盘中,根据行号+列号的奇偶性可以分成一个二分图)
对于两个相邻的合法小方格,他们必定一个属于左边的点集,一个属于右边的点集. 那么在左右点集之间连接一条边.
最终我们要求的就是该图的最大匹配边。
输出任意一组解,我们只需要保存二分图节点的原始坐标即可.
#include<cstdio>#include<cstring>#include<vector>using namespace std;const int maxn=1000+5;struct Max_Match{ int n,m; vector<int> g[maxn]; bool vis[maxn]; int left[maxn]; void init(int n,int m) { this->n=n;this->m=m; for(int i=1; i<=n; ++i) g[i].clear(); memset(left,-1,sizeof(left)); } bool match(int u) { for(int i=0;i<g[u].size();++i) { int v=g[u][i]; if(!vis[v]) { vis[v]=true; if(left[v]==-1 || match(left[v])) { left[v]=u; return true; } } } return false; } int solve() { int ans=0; for(int i=1; i<=n; ++i) { memset(vis,0,sizeof(vis)); if(match(i)) ++ans; } return ans; }}MM;int cas=1;int mapp[maxn][maxn];int n,m;struct Node{int x,y;Node(){}Node(int x,int y):x(x),y(y){}}node1[maxn],node2[maxn];bool check(int i,int j){if (node1[i].x+1==node2[j].x && node1[i].y==node2[j].y)return true;if (node1[i].x-1==node2[j].x && node1[i].y==node2[j].y)return true;if (node1[i].x==node2[j].x && node1[i].y+1==node2[j].y)return true;if (node1[i].x==node2[j].x && node1[i].y-1==node2[j].y)return true;return false;}int main(){int k; while(scanf("%d%d",&n,&m)!=EOF && n) { scanf("%d",&k);memset(mapp,0,sizeof(mapp)); while (k--){int r,c;scanf("%d%d",&r,&c);mapp[r][c]=-1;}int num1=0,num2=0; for (int i = 1;i<=n;i++)for (int j = 1;j<=m;j++)if (mapp[i][j]==0){if ((i+j)%2==0)node1[++num1]=Node(i,j);elsenode2[++num2]=Node(i,j);}MM.init(num1,num2);for (int i = 1;i<=num1;i++)for (int j = 1;j<=num2;j++) if (check(i,j)){MM.g[i].push_back(j);} printf("%d\n",MM.solve());for (int v = 1;v<=num2;v++){int u = MM.left[v];if (u!=-1){printf("(%d,%d)--(%d,%d)\n",node1[u].x,node1[u].y,node2[v].x,node2[v].y);}}printf("\n"); } return 0;}
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