BNUOJ 52303 Floyd-Warshall Lca+bfs最短路

题目链接：

https://www.bnuoj.com/v3/problem_show.php?pid=52303

Floyd-Warshall

Time Limit: 60000ms
Memory Limit: 1048576KB #### 问题描述 > In ICPCCamp, there are n cities and m (bidirectional) roads between cities. The i-th road is between the > ai-th city and the bi-th city. There may be roads connecting a citie to itself and multiple roads between > the same pair of cities. > Bobo has q travel plans. The i-th plan is to travel from the ui-th city to the vi-th city. He would like > to know the smallest number of roads needed to travel for each plan. It is guaranteed that cities are > connected.

输入

The first line contains 3 integers n, m, q (1 ≤ n ≤ 105
, 0 < m − n < 100, 1 ≤ q ≤ 105
).
The i-th of the following m lines contains 2 integers ai
, bi (1 ≤ ai
, bi ≤ n).
The i-th of the last q lines contains 2 integers ui
, vi (1 ≤ ui
, vi ≤ n).

输出

n lines with integers l1, l2, . . . , ln. li denotes the smallest number of roads travelling from city ui to city
vi
.

样例输入

4 5 3
1 2
1 3
1 4
2 3
3 4
2 2
2 3
2 4

样例输出

0
1
2

题意

给你n个点，m条无相边，q个询问，查询u到v的最短路。

题解

首先n有10^5，所以n2的bfs暴力每个源点是不可能的。
那么这题有什么特点呢？m-n<100，明显是张稀疏图！
如果题目给的是一颗树，那么我们就可以用logn的算法处理一个询问。（倍增LCA）
因此我们可以考虑把图拆出一颗树（其实是森林），吧所有非树边的点都存起来，对它们跑bfs求到所有其他点的最短路。
对于一个查询，首先我们先用logn求只经过树边的最短路，然后暴力枚举会进过的一个非树边上的顶点x,用d(u,x)+d(x,v)来更新答案。 这样就处理完了。

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#include<sstream>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

//数据需要开大点，否则会出现tle、re的情况。。。
const int maxn=101010;
const int maxm=18;

struct Edge {
    int u,v,ne,flag;
    Edge(int u,int v,int ne):u(u),v(v),ne(ne) {
        flag=0;
    }
    Edge() {}
} egs[maxn*2];

int head[maxn],tot;
int used[maxn];

void addEdge(int u,int v) {
    egs[tot]=Edge(u,v,head[u]);
    head[u]=tot++;
}

//dep：深度，fa:父亲，anc：祖先
int dep[maxn],fa[maxn],vis[maxn];
int anc[maxn][maxm];
void dfs(int u,int f,int d) {
    vis[u]=1;
    dep[u]=d;
    fa[u]=f;

    anc[u][0]=f;
    for(int i=1; i<maxm; i++) {
        anc[u][i]=anc[anc[u][i-1]][i-1];
    }

    for(int p=head[u]; p!=-1; p=egs[p].ne) {
        Edge& e=egs[p];
        if(e.v==f) continue;
        if(!vis[e.v]) {
            dfs(e.v,u,d+1);
        } else {
            used[u]=used[e.v]=1;
        }
    }
}

//dis:非树边上的点到所有点的距离
int dis[222][maxn];
//ha:存非树边上的点
int ha[222],last;
int myque[maxn],frt,rear;
void bfs(int id) {
    int s=ha[id];
    frt=0,rear=0;
    dis[id][s]=0;
    myque[rear++]=s;
    clr(vis,0);
    vis[s]=1;
    while(frt<rear) {
        int u=myque[frt++];
        for(int p=head[u]; p!=-1; p=egs[p].ne) {
            Edge& e=egs[p];
            if(vis[e.v]) continue;
            vis[e.v]=1;
            dis[id][e.v]=dis[id][u]+1;
            myque[rear++]=e.v;
        }
    }
}

int Lca(int u,int v) {
    if(dep[u]<dep[v]) swap(u,v);
    for(int i=maxm-1; i>=0; i--) {
        if(dep[anc[u][i]]>=dep[v]) {
            u=anc[u][i];
        }
    }
    if(u==v) return u;
    for(int i=maxm-1; i>=0; i--) {
        if(anc[u][i]!=anc[v][i]) {
            u=anc[u][i];
            v=anc[v][i];
        }
    }
    return anc[u][0];
}

int n,m,q;

int main() {
    scf("%d%d%d",&n,&m,&q);
    clr(used,0);
    clr(head,-1);
    tot=0;
    //建图
    for(int i=0; i<m; i++) {
        int u,v;
        scf("%d%d",&u,&v);
        if(u==v) continue;
        addEdge(u,v);
        addEdge(v,u);
    }

    //处理出树边、非树边
    clr(vis,0);
    clr(anc,0);
    for(int i=1; i<=n; i++) {
        if(!vis[i]) {
            dfs(i,0,1);
        }
    }

    //把非树边上的点存起来
    last=0;
    for(int i=1; i<=n; i++) if(used[i]) {
        ha[last++]=i;
    }

    //跑非树边上的点到所有点的距离
    clr(dis,0x3f);
    rep(i,0,last) bfs(i);

    while(q--) {
        int u,v;
        scf("%d%d",&u,&v);
        int lca=Lca(u,v);
        LL res=dep[u]+dep[v]-2*dep[lca];
        rep(i,0,last) {
            LL tmp=(LL)dis[i][u]+dis[i][v];
            res=min(res,tmp);
        }
        prf("%lld
",res);
    }

    return 0;
}

//end-----------------------------------------------------------------------