[Wikioi] 3 squares (maximum stream + maximum weight closed subgraph)

 

At the beginning, I thought of the pressure and saw n <= 30 give up.

Then I thought of black and white dyeing, and then I lost my brains. I didn't expect how to connect to the edge.

Simple question

After black and white dyeing, each vertex is connected to four sides, with a capacity of OO. If it is a black (white) node, the source node is connected to one edge, and the capacity is the weight. If it is a White (Black) node, connects an edge to the sink, and the capacity is the weight.

The final answer is all the lattice values and the-maximum stream (actually the smallest cut)

PS: (in fact, it is the QQ Farm that I have done before .. Http://www.cnblogs.com/iwtwiioi/p/3893519.html

Why do I do this by referring to my previous blog (http://www.cnblogs.com/iwtwiioi/p/3872099.html)

#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 read(a) a=getint()#define print(a) printf("%d", a)#define dbg(x) cout << #x << " = " << x << endl#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }inline const int max(const int &a, const int &b) { return a>b?a:b; }inline const int min(const int &a, const int &b) { return a<b?a:b; }const int N=1000, M=N<<4, oo=~0u>>1;struct ED { int u, v, next, cap; } e[M];int n, m, cnt=1;int d[N], gap[N], cur[N], p[N], ihead[N];inline int id(const int &x, const int &y) { return (x-1)*m+y; }inline void add(const int &u, const int &v, const int &c) {e[++cnt].next=ihead[u]; ihead[u]=cnt; e[cnt].v=v; e[cnt].u=u; e[cnt].cap=c;e[++cnt].next=ihead[v]; ihead[v]=cnt; e[cnt].v=u; e[cnt].u=v; e[cnt].cap=0;}int isap(const int &s, const int &t, const int &n) {int ret=0, f, u, v, i;for1(i, 0, n) cur[i]=ihead[i];gap[0]=n; u=s;while(d[s]<n) {for(i=cur[u]; i; i=e[i].next) if(e[i].cap && d[u]==d[e[i].v]+1) break;if(i) {v=e[i].v; p[v]=cur[u]=i; u=v;if(u==t) {for(f=oo; u!=s; u=e[p[u]].u) f=min(f, e[p[u]].cap);for(u=t; u!=s; u=e[p[u]].u) e[p[u]].cap-=f, e[p[u]^1].cap+=f;ret+=f;}}else {if(! (--gap[d[u]]) ) break;d[u]=n;cur[u]=ihead[u];for(i=ihead[u]; i; i=e[i].next) if(e[i].cap && d[u]>d[e[i].v]+1) d[u]=d[e[i].v]+1;++gap[d[u]];if(u!=s) u=e[p[u]].u;}}return ret;}int main() {read(n); read(m);int ans=0, t, now, S=id(n, m)+1, T=id(n, m)+2;for1(i, 1, n) for1(j, 1, m) {read(t); ans+=t;now=id(i, j);if((i+j)%2) add(S, now, t);else add(now, T, t);if(j<m) add(now, now+1, oo), add(now+1, now, oo);if(i<n) add(now, id(i+1, j), oo), add(id(i+1, j), now, oo);}print(ans-isap(S, T, T));return 0;}

 

 

 

 

 

Description Description

«Problem description:
In a checker with M * n squares, each square has a positive integer. We need to take the number from the square
The square where the number 2 is located does not have a public edge, and the sum of the retrieved numbers is the largest. Design a number fetch algorithm that meets the requirements.
«Programming task:
For a given checker, program the number to find the maximum sum.

Input description Input description

Row 1st has two positive integers m and n, indicating the number of rows on the board respectively.
And the number of columns. In the next m row, each row has n positive integers, indicating the number in the checker square.

Output description Output description

Output the maximum sum of the number.

Sample Input Sample Input

3 3
1 2 3
3 2 3
2 3 1

Sample output Sample output

11

Data range and prompt Data size & hint

N, m <= 30

 

[Wikioi] 3 squares (maximum stream + maximum weight closed subgraph)