P4645 [COCI2006-2007 Contest#3] BICIKLI

$ \color{#0066ff}{ 题目描述 }$

给定一个有向图，n个点，m条边。请问，1号点到2号点有多少条路径？如果有无限多条，输出inf，如果有限，输出答案模\(10^9\)的余数。

两点之间可能有重边，需要看成是不同的路径。

\(\color{#0066ff}{输入格式}\)

第一行两个整数n，m，表示点数和边数

接下来是m行有向边

\(\color{#0066ff}{输出格式}\)

输出一行表示答案

\(\color{#0066ff}{输入样例}\)

6 7
1 3
1 4
3 2
4 2
5 6
6 5
3 4
    
    
6 8
1 3
1 4
3 2
4 2
5 6
6 5
3 4
4 3

\(\color{#0066ff}{输出样例}\)

3
    
inf

\(\color{#0066ff}{数据范围与提示}\)

1 ≤ N ≤ 10 000, 1 ≤ M ≤ 100 000

\(\color{#0066ff}{题解}\)

路径无限？？显然是有环，于是直接tarjan缩点

然后记忆化搜索路径，注意，无解当且仅当路径中有siz>1的分量，为了保证是路径中的，你要在搜到目标的时候再判断!

#include<bits/stdc++.h>
#define LL long long
LL in() {
	char ch; LL x = 0, f = 1;
	while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
	for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));
	return x * f;
}
const int mod = 1e9;
const int maxn = 1e5 + 10;
struct node {
	int to;
	node *nxt;
	node(int to = 0, node *nxt = NULL): to(to), nxt(nxt) {}
};
node *head[maxn];
bool flag;
int dfn[maxn], low[maxn], cnt, st[maxn], top, ins[maxn], num, siz[maxn], bel[maxn], n, m;
LL f[maxn];
bool vis[maxn];
std::vector<int> mp[maxn];
std::set<int> have[maxn];
void add(int from, int to) { head[from] = new node(to, head[from]); }
void tarjan(int x) {
	dfn[x] = low[x] = ++cnt;
	ins[st[++top] = x] = true;
	for(node *i = head[x]; i; i = i->nxt) {
		if(!dfn[i->to]) tarjan(i->to), low[x] = std::min(low[x], low[i->to]);
		else if(ins[i->to]) low[x] = std::min(low[x], dfn[i->to]);
	}
	if(dfn[x] == low[x]) {	
		num++;
		do {
			ins[st[top]] = false;
			bel[st[top]] = num;
			siz[num]++;
			top--;
		}while(st[top + 1] != x);
	}
}
LL dfs(int x, bool now) {
	if(vis[x]) return f[x];
	if(x == bel[2]) {
		if(now) flag = true;
		return f[x] = 1;
	}
	vis[x] = true;
	for(auto go : mp[x]) {
		(f[x] += dfs(go, now | (siz[x] > 1))) %= mod;
		if(flag) return 0;
	}
	return f[x];
}


int main() {
	n = in(); m = in();
	int x, y;
	for(int i = 1; i <= m; i++) x = in(), y = in(), add(x, y);
	for(int i = 1; i <= n; i++) if(!dfn[i]) tarjan(i);
	for(int i = 1; i <= n; i++)
		for(node *o = head[i]; o; o = o->nxt)
			if(bel[i] != bel[o->to]) {
				if(have[bel[i]].count(bel[o->to])) continue;
				mp[bel[i]].push_back(bel[o->to]);
			}
	LL ans = dfs(bel[1], 0);
	if(flag) puts("inf");
	else printf("%lld\n", ans);
	return 0;
}