POJ 1236 Network of Schools

来源：互联网发布：mac上的解压缩软件编辑：程序博客网时间：2024/05/21 22:29

强连通分量

问题一，缩点后，求出所有入度为0的点的个数；

问题二，缩点后，求出入度为0的点的个数和出度为0的点的个数中的最大值

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <vector>using namespace std;vector<int> g[120];int tg[120];int dfn[120];int low[120];bool ins[120];int s[120],ls;int incnt[120],outcnt[120];int key;int tag;int min(int a,int b){    return a>b?b:a;}void tarjan(int u){    int v,i;    dfn[u]=key;    low[u]=key;    key++;    ins[u]=true;    s[ls]=u;    ls++;    for (i=0; i<g[u].size(); i++)    {        v=g[u][i];        if (dfn[v] == -1)        {            tarjan(v);            low[u]=min(low[u],low[v]);        }        else if (dfn[v] != -1 && ins[v] == true)        {            low[u]=min(low[u],dfn[v]);        }    }    if (dfn[u] == low[u])    {        do        {            v=s[ls-1];            ls--;            ins[v]=false;            tg[v]=tag;        }while (v != u);        tag++;    }}int main(){    int n,i,j,t1,res1,res2;    scanf("%d",&n);    for (i=1; i<=n; i++)    {        while (1)        {            scanf("%d",&t1);            if (t1 == 0)                break;            g[i].push_back(t1);        }    }    memset(dfn,-1,sizeof(dfn));    memset(ins,false,sizeof(ins));    key=1;    tag=1;    ls=0;    for (i=1; i<=n; i++)    {        if (dfn[i] == -1)            tarjan(i);    }    tag--;    memset(incnt,0,sizeof(incnt));    memset(outcnt,0,sizeof(outcnt));    for (i=1; i<=n; i++)    {        for (j=0; j<g[i].size(); j++)        {            if (tg[i] == tg[g[i][j]])                continue;            incnt[tg[g[i][j]]]++;            outcnt[tg[i]]++;        }    }    res1=0;    res2=0;    for (i=1; i<=tag; i++)    {      //  printf("%d %d\n",incnt[i],outcnt[i]);        if (incnt[i] == 0)            res1++;        if (outcnt[i] == 0)            res2++;    }    printf("%d\n",res1);    if (tag == 1)    {        printf("0\n");    }    else    {        printf("%d\n",res1>res2?res1:res2);    }}