POJ 1986 Distance Queries 離線LCA

Distance Queries

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 8694		Accepted: 3052
Case Time Limit: 1000MS

Description

Farmer John's cows refused to run in his marathon since he chose a path much too long for their leisurely lifestyle. He therefore wants to find a path of a more reasonable length. The input to this problem consists of the same input as in "Navigation Nightmare",followed by a line containing a single integer K, followed by K "distance queries". Each distance query is a line of input containing two integers, giving the numbers of two farms between which FJ is interested in computing distance (measured in the length of the roads along the path between the two farms). Please answer FJ's distance queries as quickly as possible!

Input

* Lines 1..1+M: Same format as "Navigation Nightmare"

* Line 2+M: A single integer, K. 1 <= K <= 10,000

* Lines 3+M..2+M+K: Each line corresponds to a distance query and contains the indices of two farms.

Output

* Lines 1..K: For each distance query, output on a single line an integer giving the appropriate distance.

Sample Input

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

Sample Output

13
3
36

Hint

Farms 2 and 6 are 20+3+13=36 apart.

Source

USACO 2004 February

 

題目大意：給你一顆樹，樹邊有長度，再給你多個詢問，每次詢問須要回答兩點間的最小長度。

 

題目分析：設樹根爲root，則兩點a，b間的最小長度等於dis（root，a） + dis（root，b） - 2 * dis（root，lca（a，b））。用一遍Tarjan的LCA算法求一下便可。

 

代碼以下：

 

#include <stdio.h>
#include <string.h>
const int oo = 0x3f3f3f3f;
const int maxE = 1000000;
const int maxN = 100005;
struct Edge{
    int n, v, d, lca;
};
Edge edge[maxE];
int p[maxN];
int Adj[maxN], l;
Edge qedge[maxE];
int qAdj[maxN], ll;
int vis[maxN];
int dis[maxN];
int ask[maxN];
int n, m, q;
int min(int a, int b){
    if(a > b) return b;
    return a;
}
int find(int x){
    return p[x] == x ? x : (p[x] = find(p[x]));
}
void addedge(int u, int v, int d){
    edge[l].v = v; edge[l].d = d; edge[l].n = Adj[u]; Adj[u] = l++;
    edge[l].v = u; edge[l].d = d; edge[l].n = Adj[v]; Adj[v] = l++;
}
void qaddedge(int u, int v){
    qedge[ll].v = v; qedge[ll].n = qAdj[u]; qAdj[u] = ll++;
    qedge[ll].v = u; qedge[ll].n = qAdj[v]; qAdj[v] = ll++;
}
void init(){
    for(int i = 0; i <= n; ++i) p[i] = i;
    memset(vis, 0, sizeof vis);
    memset(dis, oo, sizeof dis);
    memset(Adj, -1, sizeof Adj);
    memset(qAdj, -1, sizeof qAdj);
    l = 0;
    ll = 0;
}
int LCA(int u){
    p[u] = u;
    vis[u] = 1;
    for(int i = Adj[u]; ~i; i = edge[i].n){
        int v = edge[i].v;
        if(!vis[v]){
            dis[v] = min(dis[v], dis[u] + edge[i].d);
            LCA(v);
            p[v] = u;
        }
    }
    for(int i = qAdj[u]; ~i; i = qedge[i].n){
        int v = qedge[i].v;
        if(vis[v]){
            qedge[i].lca = find(v);
            qedge[i ^ 1].lca = qedge[i].lca;
            qedge[i].d = dis[v] + dis[u] - 2 * dis[qedge[i].lca];
            qedge[i ^ 1].d = qedge[i].d;
        }
    }
    return 0;
}
void work(){
    int u, v, d;
    while(~scanf("%d%d", &n, &m)){
        init();
        for(int i = 0; i < m; ++i){
            scanf("%d%d%d%*s", &u, &v, &d);
            addedge(u, v, d);
        }
        scanf("%d", &q);
        for(int i = 0; i < q; ++i){
            scanf("%d%d", &u, &v);
            ask[i] = ll;
            qaddedge(u, v);

        }
        dis[u] = 0;
        LCA(u);
        for(int i = 0; i < q; ++i){
            printf("%d\n", qedge[ask[i]].d);
        }
    }
}
int main(){
    work();
    return 0;
}

POJ 1986