SSL 2344 洛谷 2835 信息学奥赛一本通 1383 刻录光盘#floyd,tarjan,kosaraju#

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题目

求有多少个连通块

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分析

floyd+并查集

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floyd代码

#include <cstdio>
#include <cctype>
#include <cstring>
using namespace std;
int n,m,ans,f[201],x; bool v[201],a[201][201];
int in(){
	int ans=0; char c=getchar();
	while (!isdigit(c)) c=getchar();
	while (isdigit(c)) ans=ans*10+c-48,c=getchar();
	return ans;
}
int main(){
	n=in();
	for (int i=1;i<=n;i++) while (x=in(),x) a[i][x]=1; 
	for (int k=1;k<=n;k++)
	for (int i=1;i<=n;i++)
	for (int j=1;j<=n;j++)
	a[i][j]=a[i][j]||(a[i][k]&&a[k][j]);//图的连通性
	for (int i=1;i<=n;i++) f[i]=i;
	for (int i=1;i<=n;i++)
	for (int j=1;j<=n;j++)
	f[j]=(a[i][j])?f[i]:f[j];//连通吗
	for (int i=1;i<=n;i++) if (f[i]==i) ans++;//并查集 
	printf("%d",ans);
	return 0;
}

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分析

kosaraju

#include <cstdio>
#include <cstring>
using namespace std;
int n,m,ans,f[201],x; bool v[201],a[201][201];
void dfs1(int x){
    v[x]=1;
    for (int i=1;i<=n;i++)
    if (a[x][i]&&!v[i]) dfs1(i);
    f[++m]=x;//记录退出的顺序
}
void dfs2(int x){
    v[x]=1;
    for (int i=1;i<=n;i++)
    if (a[x][i]&&!v[i]) dfs2(i);
}
int main(){
    scanf("%d",&n);
    for (int i=1;i<=n;i++) while (scanf("%d",&x),x) a[i][x]=1; 
    for (int i=1;i<=n;i++) if (!v[i]) dfs1(i); memset(v,0,sizeof(v));
    for (int i=m;i>=1;i--) if (!v[f[i]]) ans++,dfs2(f[i]);//拓扑排序过
    printf("%d",ans);
    return 0;
}

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分析

tarjan

#include <cstdio>
#include <cstring>
using namespace std;
struct node{
	int x,y,next;
}e[20363]; 
int n,m,ans,f[201],rd[201],g,k,ls[201],low[201];
int color[201],tim,cnum,dfn[201],sta[201],top; bool v[201];
void tarjan(int x){
	dfn[x]=low[x]=++tim;
	sta[++top]=x; v[x]=1;
	int t=ls[x];
	while (t){
		if (!dfn[e[t].y]){//没有被访问过
			tarjan(e[t].y);
			low[x]=(low[x]>low[e[t].y])?low[e[t].y]:low[x];
		}
		else if (v[e[t].y]) //走过
		low[x]=(low[x]>low[e[t].y])?low[e[t].y]:low[x];
		t=e[t].next;
	}
	if (dfn[x]==low[x]){//形成连通图
		v[x]=0;
		color[x]=++cnum;
		do{
			v[sta[top]]=0;
			color[sta[top]]=cnum;
		}
		while(sta[top--]!=x);//出栈
	}
}
int main(){
    scanf("%d",&n);
    for (int i=1;i<=n;i++) 
	while (scanf("%d",&g),g) 
	e[++k].x=i,e[k].y=g,e[k].next=ls[e[k].x],ls[e[k].x]=k; //邻接表
    for (int i=1;i<=n;i++) if (!dfn[i]) tarjan(i); //没有被访问过
    for (int i=1;i<=k;i++) if (color[e[i].x]!=color[e[i].y]) rd[color[e[i].y]]++;//不经过
    for (int i=1;i<=cnum;i++) if (!rd[i]) ans++;//入度为0
    printf("%d",ans);
    return 0;
}