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nyoj 120 校园网络

2014-05-22 10:39 323 查看
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 10005

struct node
{
int x,y;
int next;
}edge[2*M];
int idx,head
;
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
;
int dfn
,low
;
int dfs_clock;
int countt;
int belong
;
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
,chu_du
;

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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