POJ 3422 Kaka's Matrix Travels（拆点+最大费用流）题解

题意：小A从左上角走到右下角，每个格子都有一个价值，经过这个格子就把价值拿走，每次只能往下或往右走，问你走k次最多能拿多少价值的东西。

思路：这里有一个限制条件就是经过之后要把东西拿走，也就是每一格的价值只能拿一次，这也能用拆点。我们把一个点拆成两个，建两条边，一条流量1费用-cost，另一条流量k-1，费用0，这样就完成了。

代码：

#include<cstdio>
#include<vector>
#include<stack>
#include<queue>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#define ll long long
const int maxn = 20000+5;
const int maxm = 100000+5;
const int MOD = 1e7;
const int INF = 0x3f3f3f3f;
using namespace std;
struct Edge{
    int to,next,cap,flow,cost;
}edge[maxm];
int head[maxn],tot;
int pre[maxn],dis[maxn];
bool vis[maxn];
int N,M;
void init(){
    N = maxn;
    tot = 0;
    memset(head,-1,sizeof(head));
}
void addEdge(int u,int v,int cap,int cost){
    edge[tot].to = v;
    edge[tot].cap = cap;    //容量
    edge[tot].flow = 0;
    edge[tot].cost = cost;
    edge[tot].next = head[u];
    head[u] = tot++;

    edge[tot].to = u;
    edge[tot].cap = 0;
    edge[tot].flow = 0;
    edge[tot].cost = -cost;
    edge[tot].next = head[v];
    head[v] = tot++;
}
bool spfa(int s,int t){
    queue<int> q;
    for(int i = 0;i < N;i++){
        dis[i] = INF;
        vis[i] = false;
        pre[i] = -1;
    }
    dis[s] = 0;
    vis[s] = true;
    q.push(s);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        vis[u] = false;
        for(int i = head[u];i != -1;i = edge[i].next){
            int v = edge[i].to;
            if(edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i].cost){
                dis[v] = dis[u] + edge[i].cost;
                pre[v] = i;
                if(!vis[v]){
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
    return pre[t] != -1;
}

int MCMF(int s,int t,int &cost){
    int flow = 0;
    cost = 0;
    while(spfa(s,t)){
        int MIN = INF;
        for(int i = pre[t];i != -1;i = pre[edge[i^1].to]){
            if(MIN > edge[i].cap - edge[i].flow){
                MIN = edge[i].cap - edge[i].flow;
            }
        }
        for(int i = pre[t];i != -1; i = pre[edge[i^1]. to]){
            edge[i]. flow += MIN;
            edge[i^1]. flow -= MIN;
            cost += edge[i]. cost * MIN;
        }
        flow += MIN;
    }
    return flow;
}
int mp[55][55],n;
int lp(int i,int j){ return n*(i - 1) + j; };
int rp(int i,int j){ return n*n + n*(i - 1) + j; }
int main(){
    int k;
    while(scanf("%d%d",&n,&k) != EOF){
        init();
        for(int i = 1;i <= n;i++){
            for(int j = 1;j <= n;j++){
                scanf("%d",&mp[i][j]);
                addEdge(lp(i,j),rp(i,j),1,-mp[i][j]);
                addEdge(lp(i,j),rp(i,j),k - 1,0);
                if(i < n)
                    addEdge(rp(i,j),lp(i + 1,j),k,0);
                if(j < n)
                    addEdge(rp(i,j),lp(i,j + 1),k,0);
            }
        }
        addEdge(0,lp(1,1),k,0);
        addEdge(rp(n,n),rp(n,n) + 1,k,0);
        int cost;
        MCMF(0,rp(n,n) + 1,cost);
        printf("%d\n",-cost);
    }
    return 0;
}