POJ 2762 Going from u to v or from v to u?(强联通 + TopSort)

题目大意:

为了锻炼自己的儿子 Jiajia 和Wind 把自己的儿子带入到一个洞穴内,洞穴有n个房间,洞穴的道路是单向的。

每一次Wind 选择两个房间  x 和 y,   让他的儿子从一个房间走到另一个房间,(要么从 x->y  或者 y->x), Wind承诺这个是一定可以走到的。但是他不知道如何判断这个 xy一定是互通的,现在给你一个洞穴,问随机给你两个洞穴的编号,是否是相通的。

题目分析:题目意思有点偏移,其实意思是(只要能从x->y  或者 y->x 这个都算是通的), 求一下TopSort,判断能否直接连成一串。

(集训前恶补了一下数据结构,再写TopSort的时候拿以前笔记,看一下就懂了了。)

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <vector>
usingnamespace std;
#define INF 0x7ffffff
#define maxn 1005
typedef longlong LL;
#define Min(a,b) (a<b?a:b)
#define MOD 1000000007
int m, n, Time, top, ans;
int Stack[maxn], dfn[maxn], low[maxn], P[maxn][maxn], blocks[maxn], In[maxn];
bool InStack[maxn];
vector<vector<int> > G;
void init()
{
    memset(In, 0, sizeof(In));
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(P, 0, sizeof(P));
    ans = Time = top = 0;
    G.clear();
    G.resize(n+2);
}
bool TopSort()
{
    int cnt = 0;
    top = 0;
    for(int i=0; i<ans; i++)
    {
        if(In[i] == 0)
        {
            Stack[top ++] = i;
            cnt ++;
            break;
        }
    }

    while(top)
    {
        int v = Stack[--top];
        for(int i=0; i<ans; i++)
        {
            if(P[v][i] && ! --In[i])
            {
                Stack[top++] = i;
                cnt ++;
            }
        }
    }
    return cnt == ans;
}

void Tarjan(int u)
{
    dfn[u] = low[u] = ++Time;
    Stack[top++] = u;
    InStack[u] = true;
    int len = G[u].size(), v;

    for(int i=0; i<len; i++)
    {
        v = G[u][i];
        if( !low[v] )
        {
            Tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        elseif( InStack[v] )
            low[u] = min(low[u], dfn[v]);
    }
    if(low[u] == dfn[u])
    {
        do
        {
            v  = Stack[--top];
            InStack[v] = false;
            blocks[v] = ans;
        }
        while(u != v);
        ans ++;
    }
}

void solve()
{
    for(int i=1; i<=n; i++)
    {
        if(!low[i])
            Tarjan(i);
    }

    for(int i=1; i<=n; i++)
    {
        int len = G[i].size(), v;
        for(int j=0; j<len; j++)
        {
            v = G[i][j];
            if(blocks[i] != blocks[v])
            {
                int a = blocks[i], b = blocks[v];
                P[a][b] ++;
                if(P[a][b] == 1)
                {
                    In[b] ++;
                    break;
                }

            }
        }
    }
    if( !TopSort() )
        printf("No\n");
    else
        puts("Yes");
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d %d",&n, &m);
        init();
        while(m --)
        {
            int a, b;
            scanf("%d %d",&a, &b);
            G[a].push_back(b);
        }
        solve();
    }
    return0;
}
时间: 2024-11-03 21:22:10

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