nyoj 120 校园网络

来源：互联网发布：活动目录作用知乎编辑：程序博客网时间：2024/04/28 04:39

题目链接：http://acm.nyist.net/JudgeOnline/problem.php?pid=120

思路：先将原图强连通缩点为新图，统计新图中入度，出度为0的点的个数，两者取最大值即为答案。

代码如下：

#include "stdio.h"#include "string.h"#include "stack"#include "vector"using namespace std;#define N 105#define M 10005struct node{    int x,y;    int next;}edge[2*M];int idx,head[N];void Init(){ idx=0; memset(head,-1,sizeof(head)); }inline int MIN(int a,int b){ return a<b?a:b; }inline int MAX(int a,int b){ return a>b?a:b; }void Add(int x,int y){    edge[idx].x=x;    edge[idx].y=y;    edge[idx].next=head[x];    head[x]=idx++;}bool vis[N];int dfn[N],low[N];int dfs_clock;int countt;int belong[N];stack<int> q;void DFS(int x)   //强连通缩点算法{    int i,y;    q.push(x);    vis[x] = true;    dfn[x] = low[x] = ++dfs_clock;    for(i=head[x]; i!=-1; i=edge[i].next)    {        y = edge[i].y;        if(!dfn[y])        {            DFS(y);            low[x] = MIN(low[x],low[y]);        }        else if(vis[y])            low[x] = MIN(dfn[y],low[x]);    }    if(low[x]==dfn[x])    {        int temp;        countt++;        while(1)        {            temp = q.top();            q.pop();            belong[temp] = countt;            vis[temp] = false;            if(temp==x) break;        }    }}int ru_du[N],chu_du[N];int Solve(int n){    int i;    int x,y;    dfs_clock = countt = 0;    memset(dfn,0,sizeof(dfn));    memset(vis,false,sizeof(vis));    memset(belong,0,sizeof(belong));    for(i=1; i<=n; ++i)    {        if(!dfn[i])            DFS(i);    }    if(countt==2) return 0;  //该图原本就是强连通图，return 0;    memset(ru_du,0,sizeof(ru_du));    memset(chu_du,0,sizeof(chu_du));    for(i=0; i<idx; ++i)    {        x = belong[edge[i].x];        y = belong[edge[i].y];        if(x!=y)  //统计入度，出度        {            chu_du[x]++;            ru_du[y]++;        }    }    int ru_0=0,chu_0=0;    for(i=1; i<countt; ++i)    {        if(ru_du[i]==0) ru_0++;        if(chu_du[i]==0) chu_0++;    }    int ans = MAX(ru_0,chu_0); //return 入度，出度最大的那个数    return ans;}int main(){    int T;    int n;    int i,k;    scanf("%d",&T);    while(T--)    {        Init();        scanf("%d",&n);        for(i=1; i<=n; ++i)        {            while(scanf("%d",&k),k)                Add(i,k);        }        printf("%d\n",Solve(n));    }    return 0;}

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