BZOJ 1051 & 强联通分量

题意:

  怎么说呢...这种题目有点概括不来....还是到原题面上看好了...

SOL:

  求出强联通分量然后根据分量重构图,如果只有一个点没有出边那么就输出这个点中点的数目.

  对就是这样.

  哦还有论边双与强联通的tarjan的不同...边双要记录边...无向图的边有两条要判断是不是一条...还有什么不同呢...我也不造了...看起来很像很好写就对了...

Code:

  

/*==========================================================================
# Last modified: 2016-03-13 19:24
# Filename: 1051.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 100000 
#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==================*/
struct Edge{
	int to,next;
}e[maxm],d[maxm];
int head[maxm],first[maxn];
int low[maxn],dfn[maxn],s[maxn],belong[maxn],in[maxn],sz[maxn];
bool instack[maxn];
int sume=1,scc,clo=0,top=0,n,m;
void addedge(int x,int y){
	sume++; e[sume].to=y; e[sume].next=first[x]; first[x]=sume;
}
void tarjan(int u){
	dfn[u]=low[u]=(++clo);
	s[++top]=u;
	instack[u]=true;
	for (int i=first[u];i;i=e[i].next){
		int y=e[i].to;
		if (!dfn[y]){
			tarjan(y);
			low[u]=min(low[u],low[y]);
		}
		else if (instack[y]) low[u]=min(dfn[y],low[u]);
	}
	if (low[u]==dfn[u]){
		scc++;
		while (true){
			int v=s[top--];
			belong[v]=scc;
			instack[v]=false;
			sz[scc]++;
			if (v==u) break;
		}
	}
}
void rebuild(){
	int cnt=0;
	FORP(i,1,n){
		for (int j=first[i];j;j=e[j].next){
			int v=e[j].to;
			if (belong[i]!=belong[v]){
				//----------addedge
				cnt++;
				d[cnt].to=belong[v];
				d[cnt].next=head[belong[i]];
				head[belong[i]]=cnt;
				//---------------------*/
			}
		}
	}
}
int main(){ 
	read(n); read(m);
	FORP(i,1,m){
		int x,y;
		read(x); read(y);
		addedge(x,y);
	}
	memset(dfn,0,sizeof(dfn));
	FORP(i,1,n) if (!dfn[i]) tarjan(i);
	rebuild();
	int ans=0;
	FORP(i,1,scc) if (!head[i]) {
		if (ans) {printf("%d",0); return 0;}
		else ans=sz[i];
	}
	printf("%d",ans);
	return 0;
}

 

posted @ 2016-03-13 20:26  YCuangWhen  阅读(...)  评论(... 编辑 收藏