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-;
 const int mod = 1e9+;
 const int inf = ;
 const int dir[][] = { {, }, {, }, {, -}, {, } };
 const int maxn = 2e5+;
 int num, head[maxn*], 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*];
 int spfa() {
     int st, ed;
     st = ed = ;
     mem2(dis);
     ++cnt;
     dis[s] = ;
     cap[s] = inf;
     cur[s] = -;
     q[ed++] = s;
     while(st<ed) {
         int u = q[st++];
         vis[u] = cnt-;
         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 ;
     cost += dis[t]*cap[t];
     flow += cap[t];
     for(int i = cur[t]; ~i; i = cur[e[i^].to]) {
         e[i].c -= cap[t];
         e[i^].c += cap[t];
     }
     return ;
 }
 int mcmf() {
     flow = cost = ;
     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], , val); head[v] = num++;
 }
 void input() {
     int x;
     for(int i = ; i<n; i++) {
         for(int j = ; j<n; j++) {
             scanf("%d", &x);
             add(i*n+j, i*n+j+nn, , x);
             add(i*n+j, i*n+j+nn, k-, );
         }
     }
     add(s, , k, );
     add(*nn-, t, k, );
     for(int i = ; i<n; i++) {
         for(int j = ; j<n; j++) {
             for(int k1 = ; k1<; k1++) {
                 int x = dir[k1][]+i;
                 int y = dir[k1][]+j;
                 if(x>=&&x<n&&y>=&&y<n) {
                     add(i*n+j+nn, x*n+y, k, );
                 }
             }
         }
     }
 }
 void init() {
     mem1(head);
     num = cnt = ;
     mem(vis);
 }
 int main()
 {
     while(~scanf("%d%d", &n, &k)) {
         init();
         nn = n*n;
         s = *nn, t = s+;
         input();
         int ans = mcmf();
         printf("%d\n", -ans);
     }
     return ;
 }