JZOJ 5305. 【NOIP2017提高A组模拟8.18】C (Standard IO)

5305. 【NOIP2017提高A组模拟8.18】C (Standard IO)

Time Limits: 1000 ms Memory Limits: 131072 KB

Description

Input

Output

Sample Input

10 11

1 2

2 3

3 4

1 4

3 5

5 6

8 6

8 7

7 6

7 9

9 10

6

1 2

3 5

6 9

9 2

9 3

9 10

Sample Output

2

2

2

4

4

1

Data Constraint

Hint

题解

trajan裸题

加个树上RMQ

用lca优化一下就好了

代码

#include<cstdio>
#include<vector>
#define N 100010
#define mo 1000000007
#define ll long long
using namespace std;

vector<long>map[N],bian[N];
long low[N],dfn[N],tot,sta[N],w[N],cnt[N],top,num;
bool b[N],c[N],in[N];

void dfs(long now)
{   long i,to;
    low[now]=dfn[now]=++tot;
    b[now]=in[now]=true;
    sta[++top]=now;
    for(i=0;i<map[now].size();i++){
        to=map[now][i];
        if(!b[to]){
            c[bian[now][i]]=true;
            dfs(to);
            low[now]=min(low[now],low[to]);
        }
        else if(in[to]&&!c[bian[now][i]])
            low[now]=min(low[now],dfn[to]);
    }
    if(low[now]==dfn[now]){
        num++;
        to=0;
        while(now!=to){
            to=sta[top--];
            in[to]=false;
            w[to]=num;
            cnt[num]++;
        }
    }
}

void tarjan(long n)
{   long i;
    for(i=1;i<=n;i++)
        if(!b[i])
            dfs(i);
}

long sum[N][20],fa[N][20],dep[N];

void build(long now)
{   long i,to;
    b[now]=false;
    for(i=0;i<map[now].size();i++){
        to=map[now][i];
        if(b[to]){
            if(w[to]!=w[now]){
                if(cnt[w[now]]>1)
                    sum[w[to]][0]=1;
                fa[w[to]][0]=w[now];
                dep[w[to]]=dep[w[now]]+1;
            }
            build(to);
        }
    }
}

long lca(long x,long y)
{   long up;
    if(dep[x]>dep[y])
        swap(x,y);
    up=19;
    while(dep[y]>dep[x]){
        while(dep[y]-(1<<up)<dep[x]&&up>=0)
            up--;
        if(up<0)break;
        y=fa[y][up--];
    }
    up=19;
    while(x!=y){
        while(fa[x][up]==fa[y][up]&&up>=0)
            up--;
        if(up<0)break;
        x=fa[x][up];
        y=fa[y][up];
        up--;
    }
    if(x==y)
        return x;
    else
        return fa[x][0];
}

long suan(long x)
{   long up,ans=(cnt[x]>1)?1:0;
    up=19;
    while(dep[x]>0){
        while(dep[x]-(1<<up)<0&&up>=0)
            up--;
        if(up<0)break;
        ans=(ans+sum[x][up])%mo;
        x=fa[x][up];
    }
    return ans;
}

int main()
{   long n,m,i,j,x,y,q,ci,ans,l;
    scanf("%ld%ld",&n,&m);
    for(i=1;i<=m;i++){
        scanf("%ld%ld",&x,&y);
        map[x].push_back(y);
        map[y].push_back(x);
        bian[x].push_back(i);
        bian[y].push_back(i);
    }
    tarjan(n);
    build(1);
    n=num;
    for(j=1;(1<<j)<=n;j++)
        for(i=1;i<=n;i++){
            fa[i][j]=fa[fa[i][j-1]][j-1];
            sum[i][j]=(sum[fa[i][j-1]][j-1]+sum[i][j-1])%mo;
        }
    scanf("%ld",&q);
    for(i=1;i<=q;i++){
        scanf("%ld%ld",&x,&y);
        x=w[x];
        y=w[y];
        l=lca(x,y);
        ci=(((suan(x)+suan(y))%mo-2*suan(l)%mo)%mo+((cnt[l]>1)?1:0))%mo;
        ans=1;
        for(j=1;j<=ci;j++)
            ans=((ll)ans<<1)%mo;
        printf("%ld\n",ans);
    }
    return 0;
}