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好久没写题解了= =这次是bzoj 1051

2014-09-01 20:34 162 查看
唉= =这道题我都想到了tarjan缩点,但是没有想到最后一步啊= =我们很容易想到反向建边然后缩点,这时候我们看由多少个联通块的入度为0,如果为1个,那就输出这个块的大小,否则输出0;

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 10010;
const int maxe = 50010;

struct edge {
int t;
edge* next;
}e[maxe * 2], *head[maxn]; int ne = 0;

int n, m;

void addedge(int f, int t) {
e[ne].t = t, e[ne].next = head[f]; head[f] = e + ne ++;
}

void read() {
scanf("%d%d", &n, &m);
for(int i = 1; i <= m; ++ i) {
int u, v;
scanf("%d%d", &u, &v);
addedge(v, u);
}
}

int dfs[maxn], back[maxn];
int s[maxn], top = 0, Index = 0;
bool vis[maxn]; int num = 0;
int size[maxn];
int belong[maxn];

void tarjan(int now) {
dfs[now] = back[now] = ++ Index; vis[now] = true;
s[++ top] = now;
for(edge* p = head[now]; p; p = p-> next) {
if(!dfs[p-> t]) tarjan(p-> t), back[now] = min(back[now], back[p-> t]);
else if(vis[p-> t] && back[now] > dfs[p-> t]) {
back[now] = dfs[p-> t];
}
}
if(dfs[now] == back[now]) {
++ num;
while(1) {
belong[s[top]] = num;
size[num] ++;
vis[s[top]] = false;
if(s[top] == now) break;
top --;
}
top --;
}
}

int f[maxn]; int in[maxn];
bool have[maxn];

void sov() {
memset(vis, false, sizeof(false));
memset(size, 0, sizeof(size));
memset(have, 0, sizeof(have));
for(int i = 1; i <= n; ++ i) {
if(!dfs[i]) tarjan(i);
}
memset(in, 0, sizeof(in));
int tol = 0;
for(int i = 1; i <= n; ++ i) {
for(edge* p = head[i]; p; p = p-> next) {
if(belong[p-> t] != belong[i]) {
in[belong[p-> t]] ++;
}
}
}
int pos = 0;
for(int i = 1; i <= num; ++ i) {
if(in[i] == 0) tol ++, pos = i;
}
if(tol == 1) printf("%d\n", size[pos]);
else printf("0\n");
}

int main() {
//freopen("test.in", "r", stdin);
read(); sov();
return 0;
}
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