[poj][2553][The Bottom of a Graph]

题目：http://poj.org/problem?id=2553

题意：一张有向图G，G图中从v可达的所有点w，也都可以达到v，这样的v称为sink。按照大小全部输出，没有输出空行。

方法：tarjan

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>

#define Max(A,B) ((A)>(B)?(A):(B))
#define Min(A,B) ((A)<(B)?(A):(B))
#define clr(a,b) memset(a, b, sizeof(a))
using namespace std;

const int MAXV = 10000+10;
const int inf = 0x3f3f3f3f;

const int N = 10010;
const int M = 50010;
int e, ev[M], nxt[M], head[N];
bool instack[N];
int dfn[N], low[N], dindex, q[N], ed;
int belong[N], bcnt;
int n, m;
bool out[N];
int ans[N];
int initout[N];
void addedge(int u, int v)
{
    ev[e] = v; nxt[e] = head[u]; head[u] = e++;
}

void tarjan(int u)
{
    int i, v;
    dfn[u] = low[u] = ++dindex;
    instack[u] = 1;
    q[ed++] = u;
    for (i = head[u]; ~i; i = nxt[i])
    {
        v = ev[i];
        if (!dfn[v])
        {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if (instack[v])
        {
            low[u] = min(low[u], dfn[v]);
        }
    }
    if (dfn[u] == low[u])
    {
        do{
            belong[v = q[--ed]] = bcnt;
            instack[v] = 0;
        }while (v != u);
        bcnt++;
    }
}

void solve()
{
    ed = bcnt = dindex = 0;
    clr(dfn, 0); clr(instack, 0);
    for (int i=0; i<n; i++)
        if (!dfn[i]) tarjan(i);
}

int main()
{
    //freopen("D:/a.txt", "r", stdin);
    while (~scanf("%d", &n) && n)
    {
        e = 0;
        clr(head, -1);
        clr(initout, 0);
        scanf("%d", &m);
        for (int i=0; i<m; i++)
        {
            int u, v;
            scanf("%d%d", &u, &v);
            addedge(u-1, v-1);
            initout[u-1]++;
        }
        solve();

        clr(out, 0);
        for (int i=0; i<n; i++)
        {
            for (int j=head[i]; ~j; j=nxt[j])
            {
                if (belong[i] != belong[ev[j]])
                {
                    out[belong[i]] = true;
                }
            }
        }
        int t = 0;
        for (int i=0; i<n; i++)
        {
            if (initout[i]==0) {ans[t++] = i+1; continue;}
            if (!out[belong[i]]) ans[t++] = i+1;
        }
        printf("%d", ans[0]);
        for (int i=1; i<t; i++) {
            printf(" %d", ans[i]);
        }
        puts("");
    }
}