poj 3422 Kaka's Matrix Travels 费用流

题目链接

给一个n*n的矩阵， 从左上角出发， 走到右下角， 然后在返回左上角，这样算两次。 一共重复k次， 每个格子有值， 问能够取得的最大值是多少， 一个格子的值只能取一次， 取完后变为0。

费用流第一题， 将每个格子拆为两个点， u向u'连一条容量为1， 费用为格子的值的边， u向u'再连一条容量为k-1， 费用为0的边。u'向他右边和下边的格子连一条容量为k， 费用为0的边， 跑一遍费用流就可以。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, a, n) for(int i = a; i<n; i++)
#define ull unsigned long long
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {1, 0}, {0, 1}, {0, -1}, {0, 1} };
const int maxn = 2e5+5;
int num, head[maxn*2], s, t, n, k, nn, dis[maxn], flow, cost, cnt, cap[maxn], q[maxn], cur[maxn], vis[maxn];
struct node
{
    int to, nextt, c, w;
    node(){}
    node(int to, int nextt, int c, int w):to(to), nextt(nextt), c(c), w(w) {}
}e[maxn*2];
int spfa() {
    int st, ed;
    st = ed = 0;
    mem2(dis);
    ++cnt;
    dis[s] = 0;
    cap[s] = inf;
    cur[s] = -1;
    q[ed++] = s;
    while(st<ed) {
        int u = q[st++];
        vis[u] = cnt-1;
        for(int i = head[u]; ~i; i = e[i].nextt) {
            int v = e[i].to, c = e[i].c, w = e[i].w;
            if(c && dis[v]>dis[u]+w) {
                dis[v] = dis[u]+w;
                cap[v] = min(c, cap[u]);
                cur[v] = i;
                if(vis[v] != cnt) {
                    vis[v] = cnt;
                    q[ed++] = v;
                }
            }
        }
    }
    if(dis[t] == inf)
        return 0;
    cost += dis[t]*cap[t];
    flow += cap[t];
    for(int i = cur[t]; ~i; i = cur[e[i^1].to]) {
        e[i].c -= cap[t];
        e[i^1].c += cap[t];
    }
    return 1;
}
int mcmf() {
    flow = cost = 0;
    while(spfa())
        ;
    return cost;
}
void add(int u, int v, int c, int val) {
    e[num] = node(v, head[u], c, -val); head[u] = num++;
    e[num] = node(u, head[v], 0, val); head[v] = num++;
}
void input() {
    int x;
    for(int i = 0; i<n; i++) {
        for(int j = 0; j<n; j++) {
            scanf("%d", &x);
            add(i*n+j, i*n+j+nn, 1, x);
            add(i*n+j, i*n+j+nn, k-1, 0);
        }
    }
    add(s, 0, k, 0);
    add(2*nn-1, t, k, 0);
    for(int i = 0; i<n; i++) {
        for(int j = 0; j<n; j++) {
            for(int k1 = 0; k1<2; k1++) {
                int x = dir[k1][0]+i;
                int y = dir[k1][1]+j;
                if(x>=0&&x<n&&y>=0&&y<n) {
                    add(i*n+j+nn, x*n+y, k, 0);
                }
            }
        }
    }
}
void init() {
    mem1(head);
    num = cnt = 0;
    mem(vis);
}
int main()
{
    while(~scanf("%d%d", &n, &k)) {
        init();
        nn = n*n;
        s = 2*nn, t = s+1;
        input();
        int ans = mcmf();
        printf("%d
", -ans);
    }
    return 0;
}