POJ 3422 Kaka's Matrix Travels (最小費用流)

題意：N*N的地圖上每格都有分數，分數只能獲取一次。有人從左上方開始，每次向右或下移動一格，到右下方爲止，記爲一次環遊。問第K次環遊後累計分數的最大值？
    第一次做最小費用流，參考別人的做法，思考了好久，還是有點不太明白，還是網絡流的題目做的太少，不太熟，還得多做幾個加深理解才行。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
using namespace std;
typedef long long LL;
typedef pair<int, int> P; ///first保存最短距離，second保存定點編號
const int maxn = 50*50*2 + 7;
const int INF = 0x3f3f3f3f;
int n, k, V;

int h[maxn], dis[maxn], pa_v[maxn], pa_e[maxn];
/// 頂點的勢, 最短距離, 父親節點, 父親節點的邊
struct edge{
    int to, cap, cost, rev;///終點， 容量， 費用， 反向邊
    edge(int to, int cap, int cost, int rev) : to(to), cap(cap), cost(cost), rev(rev) {}
};
vector<edge> G[maxn];
void add_edge(int from, int to, int cap, int cost) {
    G[from].push_back(edge(to, cap, cost, G[to].size()));
    G[to].push_back(edge(from, 0, -cost, G[from].size()-1));
}

int min_cost_flow(int s, int t, int f) {
    int res = 0;
    memset(h, 0, sizeof(h));
    while(f > 0) {
        priority_queue<P, vector<P>, greater<P> > que;
        memset(dis, INF, sizeof(dis));
        dis[s] = 0;
        que.push(P(0, s));
        while(!que.empty()) {
            P p = que.top(); que.pop();
            int v = p.second;
            if(dis[v] < p.first) continue;
            for(int i = 0; i < G[v].size(); ++i) {
                edge &e = G[v][i];
                if(e.cap > 0 && dis[e.to] > dis[v] + e.cost + h[v] - h[e.to]) {
                    dis[e.to] = dis[v] + e.cost + h[v] - h[e.to];
                    pa_v[e.to] = v;
                    pa_e[e.to] = i;
                    que.push(P(dis[e.to], e.to));
                }
            }
        }
        if(dis[t] == INF) return -1;
        for(int v = 0; v < V; ++v) {
            h[v] += dis[v];
        }

        int d = f;
        for(int v = t; v != s; v = pa_v[v]) {
            d = min(d, G[pa_v[v]][pa_e[v]].cap);
        }
        f -= d;
        res += d*h[t];
        for(int v = t; v != s; v = pa_v[v]) {
            edge &e = G[pa_v[v]][pa_e[v]];
            e.cap -= d;
            G[v][e.rev].cap += d;
        }
    }
    return res;
}

int main() {
    while(~scanf("%d%d", &n, &k)) {
        V = 2*n*n+1;
        add_edge(0, 1, k, 0);
        for(int i = 0; i < n; ++i) {
            for(int j = 0; j < n; ++j) {
                int t; scanf("%d", &t);
                int num = (i*n+j)*2+1;
                add_edge(num, num+1, 1, -t);
                add_edge(num, num+1, k, 0);
                if(i+1<n) {
                    int down = ((i+1)*n+j)*2+1;
                    add_edge(num+1, down, k, 0);
                }
                if(j+1<n) {
                    int right = (i*n+j+1)*2+1;
                    add_edge(num+1, right, k, 0);
                }
            }
        }
        printf("%d\n", -min_cost_flow(0, V-1, k));
    }
}