ZOJ_3195_Design the city(LCA+tarjan)

Design the city

Submit
Status

Description

Cerror is the mayor of city HangZhou. As you may know, the traffic system of this city is so terrible, that there are traffic jams everywhere. Now, Cerror finds out that the main reason of them is the poor design of the roads distribution, and he want to
change this situation.

In order to achieve this project, he divide the city up to N regions which can be viewed as separate points. He thinks that the best design is the one that connect all region with shortest road, and he is asking you to check some of his designs.

Now, he gives you an acyclic graph representing his road design, you need to find out the shortest path to connect some group of three regions.

Input

The input contains multiple test cases! In each case, the first line contian a interger N (1 < N < 50000), indicating the number of regions, which are indexed from 0 to N-1. In each of the following N-1 lines, there are three interger Ai, Bi, Li (1 < Li
< 100) indicating there‘s a road with length Li between region Ai and region Bi. Then an interger Q (1 < Q < 70000), the number of group of regions you need to check. Then in each of the following Q lines, there are three interger Xi, Yi, Zi, indicating the
indices of the three regions to be checked.

Process to the end of file.

Output

Q lines for each test case. In each line output an interger indicating the minimum length of path to connect the three regions.

Output a blank line between each test cases.

Sample Input

4
0 1 1
0 2 1
0 3 1
2
1 2 3
0 1 2
5
0 1 1
0 2 1
1 3 1
1 4 1
2
0 1 2
1 0 3

Sample Output

3
2

2
2

题意：在一棵树上，查询u -> v -> w的最短距离。

分析：LCA+tarjan离线算法，模板题。以前经常做的是两点的最短距离，现在换成三点其实是一样的，求出两两之间的最短距离之和，然后除2即可。易得。

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=15731

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 = 50000 + 5;
const int maxv = 50000 + 5;
const int maxq = 70000 + 5;

struct Edge{
    int u,v,w;
    Edge(){}
    Edge(int u,int v,int w){
        this -> u = u;
        this -> v = v;
        this -> w = w;
    }
}edge[maxq];

struct Q{ int v,id,next; }quary[6*maxq];
struct e{ int v,dis,next; }graph[2*maxn];

int n,m,q;
int a,b,c;
int father[maxn];
bool vis[maxn];
int ans[3*maxq];
int color[maxn];
int depth[maxn];
int nume,numq;
int heade[maxn];
int headq[maxn];
int cases=0;
void init(){
    for(int i=0;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=1;i<n;i++){
        scanf("%d%d%d",&a,&b,&c);
        add_E(a,b,c);
        add_E(b,a,c);
    }
    scanf("%d",&q);
    for(int i=1;i<=q;i++){
        scanf("%d%d%d",&a,&b,&c);
        edge[i]=Edge(a,b,c);
        add_Q(a,b,(i-1)*3+1);
        add_Q(b,a,(i-1)*3+1);
        add_Q(a,c,(i-1)*3+2);
        add_Q(c,a,(i-1)*3+2);
        add_Q(b,c,(i-1)*3+3);
        add_Q(c,b,(i-1)*3+3);
    }
}

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(0);
    if(cases) printf("\n");
    for(int i=1;i<=q;i++){
        printf("%d\n",(ans[(i-1)*3+1]+ans[(i-1)*3+2]+ans[(i-1)*3+3])/2);
    }
    cases++;
}

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

版权声明：本文为博主原创文章，未经博主允许不得转载。