HDU 1507 Uncle Tom's Inherited Land*（二分匹配，输出任意一组解）

<p style="margin: 10px auto; line-height: 19.5px; font-size: 13px; font-family: Verdana, Geneva, Arial, Helvetica, sans-serif; background-color: rgb(40, 85, 126);"> 要输出任意一组解。</p><p style="margin: 10px auto; line-height: 19.5px; font-size: 13px; font-family: Verdana, Geneva, Arial, Helvetica, sans-serif; background-color: rgb(40, 85, 126);">一开始时两边都是n*m-k个点做的，答案输出一半，但是错掉了，匹配数没有问题，就是输出解会出错。</p><p style="margin: 10px auto; line-height: 19.5px; font-size: 13px; font-family: Verdana, Geneva, Arial, Helvetica, sans-serif; background-color: rgb(40, 85, 126);">后来按照奇偶分成两部分就可以了</p>

#include<stdio.h>#include<algorithm>#include<iostream>#include<string.h>#include<vector>using namespace std;const int MAXN=510;int uN,vN;int g[MAXN][MAXN];int linker[MAXN];bool used[MAXN];bool dfs(int u){    for(int v=0;v<vN;v++){        if(g[u][v]&&!used[v]){            used[v]=true;            if(linker[v]==-1||dfs(linker[v]))            {                linker[v]=u;                return true;            }        }    }    return false;}int hungary(){    int res=0;    memset(linker,-1,sizeof(linker));    for(int u=0;u<uN;u++){        memset(used,false,sizeof(used));        if(dfs(u)) res++;    }    return res;}int a[110][110];int b[100];int main(){    int n,m,k;    int u,v;    while(scanf("%d%d",&n,&m)==2){        if(n==0&&m==0) break;        scanf("%d",&k);        memset(a,0,sizeof(a));        while(k--){            scanf("%d%d",&u,&v);            u--;v--;            a[u][v]=-1;        }        int index=0;        for(int i=0;i<n;i++){            for(int j=0;j<m;j++){                if(a[i][j]!=-1)                {                    b[index]=i*m+j;                    a[i][j]=index++;                }            }        }        uN=vN=index;        memset(g,0,sizeof(g));        for(int i=0;i<n;i++)            for(int j=0;j<m;j++)            if(a[i][j]!=-1&&(i+j)%2==1)                if(a[i][j]!=-1 && (i+j)%2==1)                {                    u = a[i][j];                    if(i > 0 && a[i-1][j]!=-1)                        g[u][a[i-1][j]]=1;                    if(i < n-1 && a[i+1][j]!=-1)                        g[u][a[i+1][j]]=1;                    if(j > 0 && a[i][j-1]!=-1)                        g[u][a[i][j-1]]=1;                    if(j < m-1 && a[i][j+1]!=-1)                        g[u][a[i][j+1]]=1;                }        int ans = hungary();        printf("%d\n",ans);        for(int i = 0;i <vN;i++)            if(linker[i]!=-1)            {                int x1 = b[i]/m;                int y1 = b[i]%m;                int x2 = b[linker[i]]/m;                int y2 = b[linker[i]]%m;                printf("(%d,%d)--(%d,%d)\n",x1+1,y1+1,x2+1,y2+1);            }        printf("\n");    }    return 0;}