P4645 [COCI2006-2007 Contest#3] BICIKLI

题意翻译

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。

题目描述

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.

输入输出格式

输入格式：

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.

输出格式：

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".

输入输出样例

输入样例#1：

输出样例#1：

输入样例#2：

输出样例#2：

inf

说明

本题数据已经被更改，无需保留前导0

Solution：

　　本题Tarjan缩点+拓扑序dp。

　　考虑结果为inf的情况，只要1经过一个环再到2就说明有无数条路径。

　　于是我们建两幅图，一正一反，分别处理出1能到达的所有点和能到达2的所有点，然后再对原图缩点（小优化是只需要对1能访问的点缩点就好了），枚举每个点作为中间点，判断1能否经过该点到达2且该点在一个环上，若满足条件说明解为inf咯。

　　判完inf的情况，说明剩下的图是个DAG了，直接拓扑序dp就好了（注意模数是$1e9$，开始下意识当作$1e9+7$，然后WA了，咕咕咕～）。

代码：

/*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=,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=;char x=getchar();

    while(x<''||x>'')x=getchar();

    while(x>=''&&x<='')a=(a<<)+(a<<)+(x^),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]=;

    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]=;

    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+]!=u)

            bl[stk[top]]=scc,ins[stk[top--]]=,siz[scc]++;

    }

}

queue<int>q;

int main(){

    n=gi(),m=gi();

    For(i,,m) e[i].u=gi(),e[i].v=gi(),add(e[i].u,e[i].v);

    dfs(,vis1,h,to),dfs(,vis2,H,To);

    if(!vis1[]) cout<<,exit();

    memset(h,,sizeof(h)),cnt=;

    For(i,,m) if(vis1[e[i].u]&&vis1[e[i].v]) add(e[i].u,e[i].v),rd[e[i].v]++;

    For(i,,n) if(!dfn[i]&&vis1[i]) tarjan(i);

    if(bl[]==bl[]) cout<<"inf",exit();

    For(i,,n) if(vis1[i]&&vis2[i]&&siz[bl[i]]>) cout<<"inf",exit();

    q.push(),ans[]=;

    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[];

    return ;

}