POJ_1986_Distance Queries(LCA+tarjan)

Submit Status

7 6
1 6 13 E
6 3 9 E
3 5 7 S
4 1 3 N
2 4 20 W
4 7 2 S
3
1 6
1 4
2 6

13
3
36

题意：在一棵树上，查询(u,v)最短距离。

分析：LCA+tarjan离线算法，模板题。

ps：最近写LCA用的都是数组来存边，感觉比vector好用多了~~~~~

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=11131

代码清单：

//#pragma comment(linker, "/STACK:102400000,102400000")
#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<string>
#include<cstdio>
#include<cstring>
#include<cctype>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;

typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;

const int maxn = 40000 + 5;
const int maxv = 40000 + 5;
const int maxq = 10000 + 5;

struct MAX{
    int v,d;
    MAX(){}
    MAX(int v,int d){
        this -> v = v;
        this -> d = d;
    }
};

struct Q{ int v,id,next; }quary[2*maxq];
struct e{ int v,dis,next; }graph[2*maxn];
int n,m,q;
int a,b,c;
char s[3];
int father[maxn];
bool vis[maxn];
int ans[maxq];
int color[maxn];
int depth[maxn];
int nume,numq;
int heade[maxn];
int headq[maxn];

void init(){
    for(int i=1;i<=maxn;i++) father[i]=i;
    memset(ans,-1,sizeof(ans));
    memset(color,0,sizeof(color));
    memset(depth,0,sizeof(depth));
    memset(vis,false,sizeof(vis));
    memset(graph,0,sizeof(graph));
    memset(quary,0,sizeof(quary));
    memset(heade,-1,sizeof(heade));
    memset(headq,-1,sizeof(headq));
    nume=numq=0;
}

void add_E(int u,int v,int dis){
    graph[nume].v=v;
    graph[nume].dis=dis;
    graph[nume].next=heade[u];
    heade[u]=nume++;
}

void add_Q(int u,int v,int id){
    quary[numq].v=v;
    quary[numq].id=id;
    quary[numq].next=headq[u];
    headq[u]=numq++;
}

void input(){
    for(int i=0;i<m;i++){
        scanf("%d%d%d%s",&a,&b,&c,s);
        add_E(a,b,c);
        add_E(b,a,c);
    }
    scanf("%d",&q);
    for(int i=1;i<=q;i++){
        scanf("%d%d",&a,&b);
        add_Q(a,b,i);
        add_Q(b,a,i);
    }
}

int Find(int x){ return x!=father[x] ? father[x]=Find(father[x]) : father[x]; }

void tarjan(int u){
    color[u]=1;
    vis[u]=true;
    for(int i=headq[u];i!=-1;i=quary[i].next){
        int ID=quary[i].id;
        if(ans[ID]!=-1) continue;
        int v=quary[i].v;
        if(color[v]==0) continue;
        if(color[v]==1) ans[ID]=depth[u]-depth[v];
        if(color[v]==2) ans[ID]=depth[u]+depth[v]-2*depth[Find(v)];
    }
    for(int i=heade[u];i!=-1;i=graph[i].next){
        int vv=graph[i].v;
        int dis=graph[i].dis;
        if(!vis[vv]){
            depth[vv]=depth[u]+dis;
            tarjan(vv);
            color[vv]=2;
            father[vv]=u;
        }
    }
}

void solve(){
    tarjan(1);
    for(int i=1;i<=q;i++){
        printf("%d\n",ans[i]);
    }
}

int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        init();
        input();
        solve();
    }
    return 0;
}

POJ_1986_Distance Queries(LCA+tarjan)

原文：http://blog.csdn.net/jhgkjhg_ugtdk77/article/details/47266515