Vijos1653 疯狂的方格取数

题目链接：https://vijos.org/p/1653

描述

在一个宽M，长N的矩阵中，请你编一个程序，n次从矩阵的左上角走到矩阵的右下角，每到一处，就取走该处的数字，请你选择一
种走法使取得的数字的和最大,并输出其最大值。其中：3<=M<=20 M<=N<=100 1<=n<=10

K路方格取数的费用流做法

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#define rep(i,l,r) for(int i=l; i<=r; i++)
#define clr(x,y) memset(x,y,sizeof(x))
#define travel(x) for(Edge *p=last[x]; p; p=p->pre)
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 110;
const int maxm = 30;
inline int read(){
    int ans = 0, f = 1;
    char c = getchar();
    for(; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
    for(; isdigit(c); c = getchar())
    ans = ans * 10 + c - '0';
    return ans * f;
}
struct Edge{
    Edge *pre,*rev; int to,cap,cost;
}edge[maxn*maxm*4],*last[maxn*maxm*2],*pre[maxn*maxm*2],*pt;
int n,m,k,N,S,T,pos[maxn][maxm][2],w[maxn][maxm],d[maxn*maxm*2];
bool isin[maxn*maxm*2];
queue <int> q;
inline void add(int x,int y,int z,int w){
    pt->pre = last[x]; pt->to = y; pt->cap = z; pt->cost = w; last[x] = pt++;
    pt->pre = last[y]; pt->to = x; pt->cap = 0; pt->cost = -w; last[y] = pt++;
    last[x]->rev = last[y]; last[y]->rev = last[x];
}
bool spfa(){
    clr(isin,0); isin[S] = 1; q.push(S);
    clr(d,INF); d[S] = 0;
    while (!q.empty()){
        int now = q.front(); q.pop(); isin[now] = 0;
        travel(now){
            if (p->cap && d[p->to] > d[now] + p->cost){
                d[p->to] = d[now] + p->cost;
                pre[p->to] = p;
                if (!isin[p->to]){
                    isin[p->to] = 1; q.push(p->to);
                }
            }
        }
    }
    return d[T] != INF;
}
int mincost(){
    int ret = 0, x = INF;
    while (spfa()){
        for(Edge *p = pre[T]; p; p = pre[p->rev->to]) x = min(x,p->cap);
        for(Edge *p = pre[T]; p; p = pre[p->rev->to]){
            ret += x * p->cost; p->cap -= x; p->rev->cap += x;
        }
    }
    return ret;
}
int main(){
    k = read(); m = read(); n = read();
    clr(last,0); pt = edge;
    rep(i,1,n){
        rep(j,1,m){
            w[i][j] = read();
            pos[i][j][0] = ++N; pos[i][j][1] = ++N;
            add(pos[i][j][0],pos[i][j][1],1,-w[i][j]);
            add(pos[i][j][0],pos[i][j][1],INF,0);
        }
    }
    S = ++N; T = ++N;
    add(S,pos[1][1][0],k,0); add(pos[n][m][1],T,k,0);
    rep(i,1,n) rep(j,1,m){
        if (i + 1 <= n) add(pos[i][j][1],pos[i+1][j][0],INF,0);
        if (j + 1 <= m) add(pos[i][j][1],pos[i][j+1][0],INF,0);
    }
    printf("%d
",-mincost());
    return 0;
}