HDU2225奔小康赚大钱（KM匹配模板题）

题意：略。
入门KM的模板题目，学习链接

#include<cstdio>#include<iostream>#include<algorithm>#include<cmath>#include<set>#include<map>#include<string>#include<cstring>#include<stack>#include<queue>#include<vector>#include<cstdlib>#define lson (rt<<1),L,M#define rson (rt<<1|1),M+1,R#define M ((L+R)>>1)#define cl(a,b) memset(a,b,sizeof(a));#define LL long long#define P pair<int,int>#define X first#define Y second#define pb push_back#define fread(zcc)  freopen(zcc,"r",stdin)#define fwrite(zcc) freopen(zcc,"w",stdout)using namespace std;const int maxn=305;const int inf=999999;/*   KM算法 *   复杂度O(nx*nx*ny) *   求最大权匹配 *   若求最小权匹配，可将权值取相反数，结果取相反数 *   点的编号可以从0开始，记得修改对应的循环起止 */int w[maxn][maxn];int linker[maxn],lx[maxn],ly[maxn],slack[maxn];bool visx[maxn],visy[maxn];int nx,ny;bool dfs(int x){    visx[x]=true;    for(int y=1;y<=ny;y++){        if(visy[y])continue;        int tmp=lx[x]+ly[y]-w[x][y];        if(tmp==0){            visy[y]=true;            if(linker[y]==-1||dfs(linker[y])){                linker[y]=x;                return true;            }        }else if(slack[y]>tmp){            slack[y]=tmp;        }    }    return false;}int km(){    cl(linker,-1);    cl(ly,0);    for(int i=1;i<=nx;i++){        lx[i]=-inf;        for(int j=1;j<=ny;j++)if(w[i][j]>lx[i]){            lx[i]=w[i][j];        }    }    for(int x=1;x<=nx;x++){        for(int i=1;i<=ny;i++){            slack[i]=inf;        }        while(true){            cl(visx,false);            cl(visy,false);            if(dfs(x))break;            int d=inf;            for(int i=1;i<=ny;i++)if(!visy[i]&&d>slack[i]){                d=slack[i];            }            for(int i=1;i<=nx;i++)if(visx[i]){                lx[i]-=d;            }            for(int i=1;i<=ny;i++)if(visy[i])ly[i]+=d;            else slack[i]-=d;        }    }    int ans=0;    for(int i=1;i<=ny;i++)if(linker[i]!=-1){        ans+=w[linker[i]][i];    }    return ans;}int main(){    int n;    while(~scanf("%d",&n)){        for(int i=1;i<=n;i++){            for(int j=1;j<=n;j++){                scanf("%d",&w[i][j]);            }        }        nx=ny=n;        printf("%d\n",km());    }    return 0;}