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2 years ago
#include <bits/stdc++.h>
using namespace std;
const int N = 5005; //生物种类上限
const int M = 500005; //吃与被吃的关系数上限
const int MOD = 80112002; //最大食物链数量模
int n; //生物种类
int m; //吃与被吃的关系数
int ans; //为最大食物链数量模上 80112002 的结果
vector<int> edge[N]; //保存DAG图的邻接表
queue<int> q; //广搜的队列
int f[N]; //每个生物种类的食物链最长值
int ind[N]; //每个生物种类的入度
int main() {
cin >> n >> m;
//m种关系
for (int i = 1; i <= m; i++) {
int x, y;
cin >> x >> y; //x被y吃掉
ind[y]++; //点y的入度+1
edge[x].push_back(y); //用邻接表记录下食物链的关系,x被y吃掉,由x向y引一条有向边
}
//找到所有入度为0的点放入广度优先搜索的队列
for (int i = 1; i <= n; i++) if (ind[i] == 0) q.push(i), f[i] = 1;
//f[i]=1base case,它到它的每个孩子都有一条出边,就是一条路径
//广度优先搜索DAG就是拓扑排序的模板
while (!q.empty()) {
int x = q.front();
q.pop();
for (int i = 0; i < edge[x].size(); i++) { //遍历所有出边
int y = edge[x][i]; //目标结点
f[y] = (f[x] + f[y]) % MOD; //在计算f[y]之前f[x]都是计算过的了,累加
//对接点入度-1抹去这条入边
ind[y]--;
//如果入度为0则入队列准备处理它
if (!ind[y]) q.push(y);
}
}
//遍历所有结点如果出度为0描述是食物链的顶端生物
for (int i = 1; i <= n; i++) if (edge[i].size() == 0) ans = (ans + f[i]) % MOD;
//输出大吉
cout << ans << endl;
return 0;
}