BZOJ 1412 & 最小割

什么时候ZJ省选再现一次这么良心的题吧...

题意:

  在一个染色的格子画分割线,使其不想连,求最少的线段

SOL:

  裸裸的最小割.题目要求两种颜色不想连,我们把他分到两个集合,也就是把所有相连的边切断-----这不就是最小割嘛. 把其中一个颜色与源相连,另一个颜色与汇相连,容量为正无穷,然后中间相连的容量均为1,然后跑下dinic即可.

Code:

  

/*==========================================================================
# Last modified: 2016-03-11 18:09
# Filename: 1412.cpp
# Description: 
==========================================================================*/
#define me AcrossTheSky 
#include <cstdio> 
#include <cmath> 
#include <ctime> 
#include <string> 
#include <cstring> 
#include <cstdlib> 
#include <iostream> 
#include <algorithm> 
  
#include <set> 
#include <map> 
#include <stack> 
#include <queue> 
#include <vector> 
 
#define lowbit(x) (x)&(-x) 
#define FOR(i,a,b) for((i)=(a);(i)<=(b);(i)++) 
#define FORP(i,a,b) for(int i=(a);i<=(b);i++) 
#define FORM(i,a,b) for(int i=(a);i>=(b);i--) 
#define ls(a,b) (((a)+(b)) << 1) 
#define rs(a,b) (((a)+(b)) >> 1) 
#define getlc(a) ch[(a)][0] 
#define getrc(a) ch[(a)][1] 
 
#define maxn 10005 
#define maxm 100000 
#define pi 3.1415926535898 
#define _e 2.718281828459 
#define inf 1070000000 
using namespace std; 
typedef long long ll; 
typedef unsigned long long ull; 
 
template<class T> inline 
void read(T& num) { 
    bool start=false,neg=false; 
    char c; 
    num=0; 
    while((c=getchar())!=EOF) { 
        if(c=='-') start=neg=true; 
        else if(c>='0' && c<='9') { 
            start=true; 
            num=num*10+c-'0'; 
        } else if(start) break; 
    } 
    if(neg) num=-num; 
} 
/*==================split line==================*/ 
using namespace std;
int first[maxn],d[maxn],cur[maxn];
bool vis[maxn];
int cnt=1,n,m;
int xx[4]={0,0,1,-1},yy[4]={1,-1,0,0},mp[105][105];
struct data{int to,next,v;}e[500001];
int T,S;
void ins(int u,int v,int w)
{e[++cnt].to=v;e[cnt].next=first[u];e[cnt].v=w;first[u]=cnt;}
void insert(int u,int v,int w)
{ins(u,v,w);ins(v,u,0);}

int bfs(){
	queue<int> q;
    for(int i=S;i<=T;i++) vis[i]=false;
    q.push(0); d[0]=0; vis[0]=true;
    while (!q.empty()){
        int now=q.front(); q.pop();
        for (int i=first[now];i;i=e[i].next)
            if (!vis[e[i].to] && e[i].v){
                d[e[i].to]=d[now]+1;
                vis[e[i].to]=true;
                q.push(e[i].to);
            }
    }
    return vis[T];
}
int dfs(int now,int a){
    if (now==T || !a) return a;
    int f,flow=0;
    for (int & i=cur[now];i;i=e[i].next)
        if (d[now]+1==d[e[i].to] && (f=dfs(e[i].to,min(a,e[i].v)))>0){
            flow+=f; a-=f; e[i].v-=f; e[i^1].v+=f;
            if (!a) break;
        }
    return flow;
     
}
int dinic(){
	int ans=0;
	while(bfs()){
		FORP(i,0,T) cur[i]=first[i];
		ans+=dfs(0,inf);
	}
	return ans;
}
void init()
{
     read(n); read(m);
     T=n*m+1,S=0;
     FORP(i,1,n)
     	FORP(j,1,m) read(mp[i][j]);
}
void build()
{
     for(int i=1;i<=n;i++)
         for(int j=1;j<=m;j++)
         {
             if(mp[i][j]==1)insert(0,(i-1)*m+j,inf);
             else if(mp[i][j]==2)insert((i-1)*m+j,T,inf);
             for(int k=0;k<4;k++)
             {
                     int nowx=i+xx[k],nowy=j+yy[k];
                     if(nowx<1||nowx>n||nowy<1||nowy>m||mp[i][j]==2)continue;
                     if(mp[i][j]!=1||mp[nowx][nowy]!=1)
                     insert((i-1)*m+j,(nowx-1)*m+nowy,1);
                     }
             }
     }
int main()
{
    init();
    build();
    printf("%d",dinic());
    return 0;
}

 

Sometimes it s the very people who no one imagines anything of. who do the things that no one can imagine.
posted @ 2016-03-11 18:34  YCuangWhen  阅读(141)  评论(0编辑  收藏  举报