POJ - 2186 - Popular Cows （tarjan）

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Popular Cows

题目传送：Popular Cows
思路：tarjan算法求强连通分量
AC代码：
#include <map>#include <set>#include <cmath>#include <deque>#include <queue>#include <stack>#include <cstdio>#include <cctype>#include <string>#include <vector>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>#define LL long long#define INF 0x7fffffffusing namespace std;const int maxn = 10005;int n, m;vector<int> mp[maxn];int dfn[maxn];int low[maxn];int vis[maxn];int in_stack[maxn];int index;int color[maxn];int col;bool color_is_ok[maxn];stack<int> s;int sum;void tarjan(int u) {//tarjan求强连通分量     dfn[u] = low[u] = ++ index;    s.push(u);    in_stack[u] = 1;    vis[u] = 1;    sum ++;    int d = mp[u].size();    for(int i = 0; i < d; i ++) {        int v = mp[u][i];        if(!vis[v]) {            tarjan(v);            low[u] = min(low[u], low[v]);        }        else if(in_stack[v]) {            low[u] = min(low[u], dfn[v]);        }    }    if(dfn[u] == low[u]) {//每求到一个强连通分量然后将这个强连通分量染色         int v;        do {            v = s.top();            s.pop();            in_stack[v] = 0;            color[v] = col;        } while(u != v);        col ++;//为了区分每个强连通分量     }}int main() {    while(scanf("%d %d", &n, &m) != EOF) {        for(int i = 0; i <= n; i ++) {            mp[i].clear();        }        int u, v;        for(int i = 0; i < m; i ++) {            scanf("%d %d", &u, &v);            mp[u].push_back(v);        }        memset(vis, 0, sizeof(vis));        memset(in_stack, 0, sizeof(in_stack));        memset(color, 0, sizeof(color));        index = 0;        col = 1;        sum = 0;        for(int i = 1; i <= n; i ++) {            if(!vis[i]) tarjan(i);        }//      cout << sum << endl;         if(sum != n) {            printf("0\n");            continue;        }//      for(int i = 1; i <= n; i ++) cout << color[i] << " ";//      cout << endl;         //标记有多少个缩点后出度为0的强连通分量         memset(color_is_ok, true, sizeof(color_is_ok));        for(int i = 1; i <= n; i ++) {            int d = mp[i].size();            for(int j = 0; j < d; j ++) {                if(color[mp[i][j]] != color[i]) {                    color_is_ok[color[i]] = false;                    break;                }            }        }//      for(int i = 1; i < col; i ++) {//          cout << color_is_ok[i] << " ";//      }//      cout << endl;        int ans_col;        int cnt = 0;        for(int i = 1; i < col; i ++) {            if(color_is_ok[i]) {                ans_col = i;                cnt ++;            }        }        if(cnt == 1) {//当且仅当只有一个强连通分量构成的缩点出度为0时有答案             int ans = 0;            for(int i = 1; i <= n; i ++) {                if(color[i] == ans_col) {                    ans ++;                }            }            printf("%d\n", ans);        }        else {            printf("0\n");        }    }    return 0;}
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