poj 2375 BFS实现强连通分量缩点

题意：奶牛可以向相邻的并且海拔不大于的位置进行移动，现在要添加缆车，可以试想从低处向高处行进，问最少添加多少个？

解法：

         奶牛的运动：向相同的点（双向），向低处（单向）-> 缩点，然后会使整个图成为有向图强连通图，Max（入度为0的个数，初度为0的个数），特别的当连通分量只有一个的时候输出0。

        根据本题的性质，强连通分量都是高度相同的连通区域，对于这些区域BFS就可以试想缩点，提高效率。

/*----------------------------------   Love is more than a word.   It says so much.   When I see these four letters,   I almost feel your touch.   This is only happened since   I fell in love with you.   Why this word does this,   I haven't got a clue.                   To My Goddess                          CY----------------------------------*/#include<iostream>#include<cstring>#include<algorithm>#include<cstdlib>#include<vector>#include<cmath>#include<stdlib.h>#include<iomanip>#include<list>#include<deque>#include<map>#include <stdio.h>#include <queue>#define maxn 550+5#define ull unsigned long long #define ll long long#define reP(i,n) for(i=1;i<=n;i++) #define REP(i,a,b) for(i=a;i<=b;i++)  #define rep(i,n) for(i=0;i<n;i++)#define cle(a) memset(a,0,sizeof(a)) #define clehead(a) rep(i,maxn)a[i]=-1/*  The time of story :  **  while(1)    {         once upon a time,         there was a mountain,         on top of which there was a temple,         in which there was an old monk and a little monk.         Old monk was telling stories inside the temple.         What was he talking about?  **  }   ÎûÎû   (*^__^*)*/#define sci(a) scanf("%d",&a) #define scd(a) scanf("%lf",&a)  #define pri(a) printf("%d",a)   #define prie(a) printf("%d\n",a)    #define prd(a)  printf("%lf",a)     #define prde(a) printf("%lf\n",a)      #define pre printf("\n")#define LL(x) x<<1 #define RR(x) x<<|1#define pb push_back#define mod 90001#define PI 3.141592657const ull INF = 1LL << 61;const int inf =   int(1e5)+10;const double eps=1e-5;using namespace std;struct node{   int u,v,w;   int next;};int w,l;int val[maxn][maxn];int scc_num;int used[maxn][maxn];int belong[maxn][maxn];int dir[4][2]={1,0,-1,0,0,1,0,-1};int in[2555000];int out[2555000];void init(){    cle(used);    scc_num=0;    cle(belong);}bool getbool(int x,int y){    if(x<0||y<0)return false;    if(x>=l||y>=w)return false;    return true;}bool cmp(int a,int b){    return a>b;}void bfs(int x,int y){    int i,j,k;    if(getbool(x,y)==false)return;    used[x][y]=1;    belong[x][y]=scc_num;    for(i=0;i<4;i++)    {        int tempx=x+dir[i][0];        int tempy=y+dir[i][1];        if(getbool(tempx,tempy)&&val[tempx][tempy]==val[x][y]&&used[tempx][tempy]==0)        {            bfs(tempx,tempy);        }    }    return ;}void doit(){    int i,j,k;    for(i=0;i<l;i++)    {        for(j=0;j<w;j++)        {            for(k=0;k<4;k++)            {                int x,y;                x=i+dir[k][0];                y=j+dir[k][1];                if(getbool(x,y)&&belong[i][j]!=belong[x][y]&&val[x][y]<=val[i][j])                {                    out[belong[x][y]]++;                    in[belong[i][j]]++;                }            }        }    }}int main(){    freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    int i,j,k;    while(cin>>w>>l)    {        init();        for(i=0;i<l;i++)        {            for(j=0;j<w;j++)            {                scanf("%d",&val[i][j]);            }        }        for(i=0;i<l;i++)        {            for(j=0;j<w;j++)            {                if(used[i][j]==0)                {                    scc_num++;                    bfs(i,j);                }            }        }        cle(in);        cle(out);        doit();        int ans1=0;        int ans2=0;        int ans=0;        for(i=1;i<=scc_num;i++)        {            if(in[i]==0)            {                ans1++;            }            if(out[i]==0)            {                ans2++;            }        }        if(scc_num==1)cout<<0<<endl;        else cout<<max(ans1,ans2)<<endl;    }    return 0;}