[Tyvj] QQ Farm (max stream + max weight closure diagram)

Http://tyvj.cn/Problem_Show.aspx? Id = 1338

The time is in the rank7 period. Isap I write so small constants...

View my other blog on the max permission closure Diagram

This is an obvious model.

First, we first dye the entire graph to make it black and white. Here we only need to link the white nodes to the upper and lower sides at the Black Point. Obviously, these nodes have the right value, what we need to require is the maximum authority, so we can do it just like my blog post.

Sum-least cut is the answer.

#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <iostream>#include <algorithm>using namespace std;#define rep(i, n) for(int i=0; i<(n); ++i)#define for1(i,a,n) for(int i=(a);i<=(n);++i)#define for2(i,a,n) for(int i=(a);i<(n);++i)#define for3(i,a,n) for(int i=(a);i>=(n);--i)#define for4(i,a,n) for(int i=(a);i>(n);--i)#define CC(i,a) memset(i,a,sizeof(i))#define max(a,b) ((a)>(b)?(a):(b))#define min(a,b) ((a)<(b)?(a):(b))#define read(a) a=getnum()#define print(a) printf("%d", a)inline int getnum() { int ret=0; char c; for(c=getchar(); c<‘0‘ || c>‘9‘; c=getchar()); for(; c>=‘0‘ && c<=‘9‘; c=getchar()) ret=ret*10+c-‘0‘; return ret; }const int N=40005, M=N*4*2, oo=~0u>>2;int ihead[N], inext[M], from[M], to[M], cap[M], cnt=1;int gap[N], d[N], cur[N], p[N];inline void add(const int &u, const int &v, const int &w) {inext[++cnt]=ihead[u]; ihead[u]=cnt; from[cnt]=u; to[cnt]=v; cap[cnt]=w;inext[++cnt]=ihead[v]; ihead[v]=cnt; from[cnt]=v; to[cnt]=u; cap[cnt]=0;}inline int isap(const int &s, const int &t, const int &n) {int u=s, f, i, flow=0;for1(i, 0, n) cur[i]=ihead[i];gap[0]=n;while(d[s]<n) {for(i=cur[u]; i; i=inext[i]) if(cap[i] && d[u]==d[to[i]]+1) break;if(i) {cur[u]=i; p[to[i]]=i; u=to[i];if(u==t) {for(f=oo; u!=s; u=from[p[u]]) f=min(f, cap[p[u]]);for(u=t; u!=s; u=from[p[u]]) cap[p[u]]-=f, cap[p[u]^1]+=f;flow+=f;}}else {if(!(--gap[d[u]])) break;d[u]=n;cur[u]=ihead[u];for(i=ihead[u]; i; i=inext[i]) if(cap[i] && d[u]>d[to[i]]+1) d[u]=d[to[i]]+1;++gap[d[u]];if(u!=s) u=from[p[u]];}}return flow;}int main() {int t, u, n;read(n);int S=0, T=n*n+1;int sum=0, last;for1(i, 1, n) {last=i;for1(j, 1, n) {read(t); sum+=t;u=(i-1)*n+j;if(last%2) {if(i>1) add(u, u-n, oo);if(i<n) add(u, u+n, oo);if(j>1) add(u, u-1, oo);if(j<n) add(u, u+1, oo);add(S, u, t);}else add(u, T, t);++last;}}printf("%d\n", sum-isap(S, T, T+1));return 0;}