这个题强联通分量容易出错的地方我都出了 我把它记下来。。。
#include <iostream>
#include <cstdio>
#include <cstring>
#define clr(x) memset(x, 0, sizeof(x))
using namespace std;
const int maxn = 50005;
const int maxm = 50005;
struct Edge {
int to, next;
}e[maxm];
int head[maxn], tot;
void add(int u, int v) {
e[tot].to = v;
e[tot].next = head[u];
head[u] = tot++;
}
int min(int a, int b) {
return a > b ? b : a;
}
int n, m;
int dfn[maxn], low[maxn], id[maxn], outd[maxn], s[maxn], ins[maxn];
int s_cnt, ans_cnt, cnt;
void init() {
clr(head);
clr(dfn); clr(low); clr(id);clr(outd); clr(ins);
tot = 1;
s_cnt = 0;
ans_cnt = 0;
cnt = 1;
}
void dfs(int u) {
s[s_cnt++] = u;
ins[u] = 1;
for(int i = head[u]; i; i = e[i].next) {
int v = e[i].to;
if(!dfn[v]) {
dfn[v] = low[v] = cnt++;
dfs(v);
low[u] = min(low[u], low[v]);
} else if(ins[v]){
low[u] = min(low[u], dfn[v]);
}
}
if(dfn[u] == low[u]) {
ans_cnt++;
int x;
do {
x = s[--s_cnt];
ins[x] = 0;
id[x] = ans_cnt;
} while(x != u);
}
}
int main() {
int x, y;
while(EOF != scanf("%d %d",&n, &m) ) {
init();
for(int i = 1; i <= m; i++) {
scanf("%d %d",&x, &y);
add(x, y);
}
for(int i = 1; i <= n; i++) {
if(!dfn[i]) {
dfn[i] = low[i] = cnt++;
dfs(i);
}
}
if(ans_cnt == 1) {
printf("%d ", n);
continue;
}
for(int i = 1; i <= n; i++) {
for(int j = head[i]; j; j = e[j].next) {
int k = e[j].to;
if(id[i] != id[k])
outd[id[i]]++;
}
}
int flag = 0;
for(int i = 1; i <= ans_cnt; i++) {
if(outd[i] == 0) {
if(flag == 0) flag = i;
else {
flag = 0;
break;
}
}
}
if(flag == 0) {
puts("0");
} else {
int sum = 0;
for(int i = 1; i <= n; i++) {
if(id[i] == flag) {
sum++;
}
}
printf("%d ", sum);
}
}
return 0;
}