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49 lines
1.6 KiB
49 lines
1.6 KiB
#include <bits/stdc++.h>
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using namespace std;
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const int N = 1010; //牧场数上限
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int n; //n个牧场
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int m; //m条有向路连接
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int K; //k只奶牛
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int ans; //ans为最终答案
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int g[N][N]; //g数组为邻接矩阵
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int a[N]; //a数组存储牛的位置
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int sum[N]; //sum数组为每个点被遍历的次数
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bool st[N]; //st数组用来判断点是否已经被访问过
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//进行图的深度优先遍历
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void dfs(int x) {
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st[x] = true; //将现在访问的点标记为已遍历,防止走回头路
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sum[x]++; //将这个点遍历的次数+1
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//枚举节点编号
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for (int i = 1; i <= n; i++)
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//如果当前节点没有被访问过并且与当前节点有边连接
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if (!st[i] && g[x][i]) dfs(i);//就遍历i号节点
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}
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int main() {
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cin >> k >> n >> m;
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for (int i = 1; i <= k; i++) cin >> a[i];//输入每只奶牛的顺序
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//使用邻接矩阵保存数据边
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for (int i = 1; i <= m; i++) {
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int u, v;
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cin >> u >> v;
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//连接两边(注意不是双向边,是单向边)
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g[u][v] = 1;
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}
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//对奶牛的位置进行枚举
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for (int i = 1; i <= k; i++) {
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dfs(a[i]); //从每一只奶牛的位置开始遍历
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memset(st, false, sizeof(st)); //记得每次遍历完都需要清空标记数组
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}
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//统计答案，如果当前节点被访问的次数恰好为奶牛的只数
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for (int i = 1; i <= n; i++)
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if (sum[i] == k) ++ans;
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//输出最后答案
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cout << ans << endl;
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return 0;//完美结束
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} |