HDU 3376 Matrix Again

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int INF = 0x3f3f3f3f;
 4 const int maxn = 601*602*2;
 5 struct arc {
 6     int to,flow,cost,next;
 7     arc(int x = 0,int y = 0,int z = 0,int nxt = -1) {
 8         to = x;
 9         flow = y;
10         cost = z;
11         next = nxt;
12     }
13 } e[3000010];
14 int head[maxn],d[maxn],p[maxn],tot,S,T;
15 bool in[maxn];
16 void add(int u,int v,int flow,int cost) {
17     e[tot] = arc(v,flow,cost,head[u]);
18     head[u] = tot++;
19     e[tot] = arc(u,0,-cost,head[v]);
20     head[v] = tot++;
21 }
22 bool spfa() {
23     memset(in,false,sizeof in);
24     memset(d,-1,sizeof d);
25     memset(p,-1,sizeof p);
26     queue<int>q;
27     d[S] = 0;
28     q.push(S);
29     while(!q.empty()) {
30         int u = q.front();
31         q.pop();
32         in[u] = false;
33         for(int i = head[u]; ~i; i = e[i].next) {
34             if(e[i].flow && d[e[i].to] < d[u] + e[i].cost) {
35                 d[e[i].to] = d[u] + e[i].cost;
36                 p[e[i].to] = i;
37                 if(!in[e[i].to]) {
38                     in[e[i].to] = true;
39                     q.push(e[i].to);
40                 }
41             }
42         }
43     }
44     return p[T] > -1;
45 }
46 int solve(int ret = 0) {
47     while(spfa()) {
48         int mh = 1;
49         //for(int i = p[T]; ~i; i = p[e[i^1].to])
50             //mh = min(mh,e[i].flow);
51         for(int i = p[T]; ~i; i = p[e[i^1].to]) {
52             e[i].flow -= mh;
53             e[i^1].flow += mh;
54         }
55         ret += mh*d[T];
56     }
57     return ret;
58 }
59 int main() {
60     int n;
61     while(~scanf("%d",&n)) {
62         memset(head,-1,sizeof head);
63         int ret = tot = 0;
64         for(int i = 1; i <= n; ++i)
65             for(int j = 1,val; j <= n; ++j) {
66                 scanf("%d",&val);
67                 int u = (i-1)*n + j,v = u + n*n;
68                 if(u == 1 || u == n*n) {
69                     add(u,v,2,val);
70                     ret -= val;
71                 } else add(u,v,1,val);
72                 if(i < n) add(v,u + n,1,0);
73                 if(j < n) add(v,u + 1,1,0);
74             }
75         S = 1;
76         T = 2*n*n;
77         printf("%d\n",solve() + ret);
78     }
79     return 0;
80 }