SGU 190 Dominoes(二分图匹配)

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题意：给出一个n×n的棋盘，现在挖掉p个格子，问剩下的空位能否用1×2的Dominoes填满，小矩形不能覆盖。输出方案。

思路：如果两个相邻的格子能放Dominoes，那么就在它们之间连一条边，很容易看出这是一个二分图，接下来只要做个最大匹配，看最大匹配数是否是空着的格子的一半就行了。

代码：

#include <iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<map>#include<queue>#include<stack>#include<cmath>#include<vector>#define inf 2139062143#define Inf 0x3FFFFFFFFFFFFFFFLL#define eps 1e-9#define pi acos(-1.0)using namespace std;typedef long long ll;const int maxn=50;const int maxm=maxn*maxn;struct Edge{    int v,next;    Edge(){};    Edge(int vv,int nx) {v=vv;next=nx;}}edges[maxm<<3];const int dir[4][2]={{1,0},{0,1},{0,-1},{-1,0}};int board[maxn][maxn],head[maxm],nEdge;int conv[maxm][2];bool vis[maxm];int links[maxm];vector<int>hor,ver;void AddEdge(int u,int v){    edges[++nEdge]=Edge(v,head[u]);    head[u]=nEdge;}void Init(){    memset(board,0,sizeof(board));    memset(head,0xff,sizeof(head));    memset(links,0xff,sizeof(links));    nEdge=-1;    hor.clear();    ver.clear();}bool dfs(int u){    for(int k=head[u];k!=-1;k=edges[k].next)    {        int v=edges[k].v;        if(!vis[v])        {            vis[v]=true;            if(links[v]==-1||dfs(links[v]))            {                links[v]=u;                return true;            }        }    }    return false;}int hungary(int n){    int ans=0;    for(int i=1;i<=n;++i)    {        for(int j=1;j<=n;++j)        {            if(board[i][j]==0)            {                memset(vis,0,sizeof(vis));                if(dfs((i-1)*n+j)) ans++;            }        }    }    return ans;}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    int n,p;    scanf("%d%d",&n,&p);    Init();    int x,y,sum=0;    for(int i=0;i<p;++i)    {        scanf("%d%d",&x,&y);        board[x][y]=-1;    }    x=0;y=0;    for(int i=1;i<=n;++i)    {        for(int j=1;j<=n;++j)        {            x^=1;            if(board[i][j]!=-1)                {board[i][j]=x;sum++;}            conv[++y][0]=i;conv[y][1]=j;        }        x^=1;    }    for(int i=1;i<=n;++i)    {        for(int j=1;j<=n;++j)        {            if(board[i][j]!=0) continue;            for(int k=0;k<4;++k)            {                x=i+dir[k][0];                y=j+dir[k][1];                if(x>=1&&x<=n&&y>=1&&y<=n&&board[x][y]!=-1)                {                    AddEdge((i-1)*n+j,(x-1)*n+y);                }            }        }    }    int ans=hungary(n);    if((sum&1)||(ans!=sum/2))    {        printf("No\n");    }    else    {        printf("Yes\n");        for(int i=1;i<=n*n;++i)        {            if(links[i]!=-1)            {                x=i;                y=links[i];                if(x>y) swap(x,y);                if(conv[x][1]==conv[y][1])                    hor.push_back(x);                else                    ver.push_back(x);            }        }        int size=hor.size();        printf("%d\n",size);        for(int i=0;i<size;++i)            printf("%d %d\n",conv[hor[i]][0],conv[hor[i]][1]);        size=ver.size();        printf("%d\n",size);        for(int i=0;i<size;++i)            printf("%d %d\n",conv[ver[i]][0],conv[ver[i]][1]);    }    return 0;}

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