Codevs[1227]方格取数2 费用流

题目链接：http://codevs.cn/problem/1227/

题目大意：传k遍纸条

每个点拆成入和出，入向出连一条容量为1，费用为该点权值的边(表示选这个点，只能走一次)
再连一条容量为INF，费用0的边(选过就不再有贡献)
每个点的出向右和下的入连一条容量为INF费用为0的边
超级源连点(1,1)的入一条容量为k，费用0的边(传k遍)，点(n,m)的出连向超级汇，容量INF，费用0

跑最大费用流

代码如下：

#include<cstring>#include<ctype.h>#include<cstdio>#include<queue>#define N 50020#define INF 2147483647using namespace std;const int S=50001;const int T=50002;inline int read(){    int x=0,f=1;char c;    do c=getchar(),f=c=='-'?-1:f; while(!isdigit(c));    do x=(x<<3)+(x<<1)+c-'0',c=getchar(); while(isdigit(c));    return x*f;}queue<int>q;int n,k,x,cost,top=1;int d[N],fir[N];bool b[N];struct Edge{    int to,nex,k,v;    Edge(int _=0,int __=0,int ___=0,int ____=0):to(_),nex(__),k(___),v(____){}}nex[250005];inline int GetNum(int x,int y){    return (x-1)*n+y;}inline bool spfa(){    for(int i=0;i<=n*n*2;i++)        d[i]=-INF,b[i]=false;    d[T]=-INF;    d[S]=0;q.push(S);    while(!q.empty()){        int x=q.front();q.pop();        b[x]=false;        for(int i=fir[x];i;i=nex[i].nex)            if(nex[i].k && d[nex[i].to]<d[x]+nex[i].v){///最大费用                d[nex[i].to]=d[x]+nex[i].v;                if(!b[nex[i].to]) b[nex[i].to]=true,q.push(nex[i].to);            }    }    return d[T]!=-INF;}int dfs(int x,int v){    if(x==T || !v){        cost=cost+v*d[T];        return v;    }    b[x]=true;    int tmp=0;    for(int i=fir[x];i;i=nex[i].nex)        if(!b[nex[i].to] && d[nex[i].to]==d[x]+nex[i].v && nex[i].k){            int f=dfs(nex[i].to,min(v,nex[i].k));            v-=f;nex[i].k-=f;nex[i^1].k+=f;tmp+=f;            if(!v) break;        }    if(!tmp) d[x]=-INF;    return tmp;}inline void Dinic(){    while(spfa()) dfs(S,INF);}inline void add(int x,int y,int k,int v){    nex[++top]=Edge(y,fir[x],k,v);    fir[x]=top;}int main(){    n=read();k=read();    for(int i=1;i<=n;i++)        for(int j=1;j<=n;j++){            x=read();            add(GetNum(i,j),GetNum(i,j)+n*n,1,x);            add(GetNum(i,j)+n*n,GetNum(i,j),0,-x);            add(GetNum(i,j),GetNum(i,j)+n*n,INF,0);            add(GetNum(i,j)+n*n,GetNum(i,j),0,0);            if(i+1<=n){                add(GetNum(i,j)+n*n,GetNum(i+1,j),INF,0);                add(GetNum(i+1,j),GetNum(i,j)+n*n,0,0);            }            if(j+1<=n){                add(GetNum(i,j)+n*n,GetNum(i,j+1),INF,0);                add(GetNum(i,j+1),GetNum(i,j)+n*n,0,0);            }        }    add(S,GetNum(1,1),k,0);add(GetNum(1,1),S,0,0);    add(GetNum(n,n)+n*n,T,INF,0);add(T,GetNum(n,n)+n*n,0,0);    Dinic();    printf("%d",cost);return 0;}