POJ2553-The Bottom of a Graph

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题目链接

题意：求解Bottom(G)，即集合内的点可以互相到达。

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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