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