POJ2553-The Bottom of a Graph

id=2553">主题链接

题意：求解Bottom(G)。即集合内的点能够互相到达。

思路：有向图的强连通。缩点，找出出度为0的点，注意符合的点要按升序输出。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAXN = 5010;
const int MAXM = 50010;

struct Edge{
    int to, next;
}edge[MAXM];

int head[MAXN], tot;
int Low[MAXN], DFN[MAXN], Stack[MAXN], Belong[MAXN];
int Index, top;
int scc;
bool Instack[MAXN];
int num[MAXN];
int n, m;
int out[MAXN], ans[MAXN];

void init() {
    tot = 0;
    memset(head, -1, sizeof(head));
}

void addedge(int u, int v) {
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;
}

void Tarjan(int u) {
    int v;
    Low[u] = DFN[u] = ++Index;
    Stack[top++] = u;
    Instack[u] = true;
    for (int i = head[u]; i != -1; i = edge[i].next) {
        v = edge[i].to;
        if (!DFN[v]) {
            Tarjan(v);
            if (Low[u] > Low[v]) Low[u] = Low[v];
        }
        else if (Instack[v] && Low[u] > DFN[v])
            Low[u] = DFN[v];
    }
    if (Low[u] == DFN[u]) {
        scc++;
        do {
            v = Stack[--top];
            Instack[v] = false;
            Belong[v] = scc;
            num[scc]++;
        } while (v != u);
    }
}

void solve() {
    memset(Low, 0, sizeof(Low));
    memset(DFN, 0, sizeof(DFN));
    memset(num, 0, sizeof(num));
    memset(Stack, 0, sizeof(Stack));
    memset(Instack, false, sizeof(Instack));
    Index = scc = top = 0;
    for (int i = 1; i <= n; i++)
        if (!DFN[i])
            Tarjan(i);
}

int main() {
    while (scanf("%d%d", &n, &m) && n) {
        init();
        int u, v;
        for (int i = 0; i < m; i++) {
            scanf("%d%d", &u, &v);
            addedge(u, v);
        }
        solve();    

        memset(out, 0, sizeof(out));
        for (int u = 1; u <= n; u++) {
            for (int i = head[u]; i != -1; i = edge[i].next) {
                int v = edge[i].to;
                if (Belong[u] != Belong[v])
                    out[Belong[u]]++;
            }
        }
        memset(ans, 0, sizeof(ans));
        int cnt = 0;
        for (int i = 1; i <= scc; i++)
            for (int u = 1; u <= n; u++) {
                if (out[i] == 0) {
                    if (Belong[u] == i)
                        ans[cnt++] = u;
                }
            }
        sort(ans, ans + cnt);
        for (int i = 0; i < cnt; i++)
            if (i == 0) printf("%d", ans[i]);
            else printf(" %d", ans[i]);
        printf("\n");
    }
    return 0;
}

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