題意翻譯

給定一個有向圖，n個點，m條邊。請問，1號點到2號點有多少條路徑？如果有無限多條，輸出inf，如果有限，輸出答案模10^9的余數。

兩點之間可能有重邊，需要看成是不同的路徑。

題目描述

A bicycle race is being organized in a land far, far away. There are N town in the land, numbered 1 through N. There are also M one-way roads between the towns. The race will start in town 1 and end in town 2. How many different ways can the route be set? Two routes are considered different if they do not use the exact same roads.

輸入輸出格式

The first line of input contains two integers N and M (1 ≤ N ≤ 10 000, 1 ≤ M ≤ 100 000), the number of towns and roads. Each of the next M lines contains two different integers A and B, representing a road between towns A and B. Towns may be connected by more than one road.

Output the number of distinct routes that can be set on a single line. If that number has more than nine digits, output only the last nine digits of the number. If there are infinitely many routes, output "inf".

輸入輸出樣例

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
 

3

inf

說明

本題數據已經被更改，無需保留前導0

Solution：

　　本題Tarjan縮點+拓撲序dp。

　　考慮結果為inf的情況，只要1經過一個環再到2就說明有無數條路徑。

　　於是我們建兩幅圖，一正一反，分別處理出1能到達的所有點和能到達2的所有點，然後再對原圖縮點（小優化是只需要對1能訪問的點縮點就好了），枚舉每個點作為中間點，判斷1能否經過該點到達2且該點在一個環上，若滿足條件說明解為inf咯。

　　判完inf的情況，說明剩下的圖是個DAG了，直接拓撲序dp就好了（註意模數是$1e9$，開始下意識當作$1e9+7$，然後WA了，咕咕咕～）。

代碼：

/*Code by 520 -- 9.5*/
#include<bits/stdc++.h>
#define il inline
#define ll long long
#define RE register
#define For(i,a,b) for(RE int (i)=(a);(i)<=(b);(i)++)
#define Bor(i,a,b) for(RE int (i)=(b);(i)>=(a);(i)--)
using namespace std;
const int N=200005,mod=1e9;
int n,m,rd[N],bl[N],scc,low[N],dfn[N],siz[N],tot;
int to[N],net[N],h[N],cnt,stk[N],top,ans[N];
int To[N],Net[N],H[N];
bool vis1[N],vis2[N],ins[N];
struct node{
    int u,v;
}e[N];
int gi(){
    int a=0;char x=getchar();
    while(x<‘0‘||x>‘9‘)x=getchar();
    while(x>=‘0‘&&x<=‘9‘)a=(a<<3)+(a<<1)+(x^48),x=getchar();
    return a;
}

il void add(int u,int v){
    to[++cnt]=v,net[cnt]=h[u],h[u]=cnt;
    To[++cnt]=u,Net[cnt]=H[v],H[v]=cnt;   
}

void dfs(int u,bool *vis,int *h,int *to){
    vis[u]=1;
    for(RE int i=h[u];i;i=net[i])
        if(!vis[to[i]]) dfs(to[i],vis,h,to);
}

void tarjan(int u){
    dfn[u]=low[u]=++tot,stk[++top]=u,ins[u]=1;
    for(RE int i=h[u];i;i=net[i])
        if(!dfn[to[i]]) tarjan(to[i]),low[u]=min(low[u],low[to[i]]);
        else if(ins[to[i]]) low[u]=min(low[u],dfn[to[i]]);
    if(dfn[u]==low[u]){
        ++scc;
        while(stk[top+1]!=u)
            bl[stk[top]]=scc,ins[stk[top--]]=0,siz[scc]++;
    }
}

queue<int>q;
int main(){
    n=gi(),m=gi();
    For(i,1,m) e[i].u=gi(),e[i].v=gi(),add(e[i].u,e[i].v);
    dfs(1,vis1,h,to),dfs(2,vis2,H,To);
    if(!vis1[2]) cout<<0,exit(0);
    memset(h,0,sizeof(h)),cnt=0;
    For(i,1,m) if(vis1[e[i].u]&&vis1[e[i].v]) add(e[i].u,e[i].v),rd[e[i].v]++;
    For(i,1,n) if(!dfn[i]&&vis1[i]) tarjan(i);
    if(bl[1]==bl[2]) cout<<"inf",exit(0);
    For(i,1,n) if(vis1[i]&&vis2[i]&&siz[bl[i]]>1) cout<<"inf",exit(0);
    q.push(1),ans[1]=1;
    while(!q.empty()){
        RE int u=q.front();q.pop();
        for(RE int i=h[u];i;i=net[i]){
            ans[to[i]]=(ans[to[i]]+ans[u])%mod;
            if(!--rd[to[i]])q.push(to[i]);
        }
    }
    cout<<ans[2];
    return 0;
}    

P4645 [COCI2006-2007 Contest#3] BICIKLI