POJ-1986 Distance Queries(LCA、离线)


Distance Queries

Time Limit: 2000MS   Memory Limit: 30000K
Total Submissions: 14378   Accepted: 5062
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

题目大意:第一行告诉你田的数量和田之间的路数量,然后告诉你两块田的距离和方向(可以忽略方向),然后k次询问,两块田间的最短距离。

解题思路:直接套模板,对于lca离线算法的理解学习,参考(http://www.cnblogs.com/JVxie/p/4854719.html),觉得很详细,就通过这个了解了离线,在线的方法至今不懂。。

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=1e5+10;

struct node
{
    int v,c;
};

vector<node>tree[maxn],que[maxn];
int dis[maxn],num[maxn],f[maxn];
bool vis[maxn];

void Init(int n)
{
    for(int i=0;i<=n;i++)
    {
        tree[i].clear();
        que[i].clear();
        f[i] = i;
        dis[i] = 0;
        num[i] = 0;
        vis[i] = 0;
    }
}

int Find(int x)
{
    int r=x;
    while(r!=f[r])
    {
        r = f[r];
    }
    while(x!=f[x])
    {
        int j=f[x];
        f[x] = r;
        x = j;
    }
    return x;
}

void lca(int u)
{
    vis[u] = true;
    f[u] = u;
    for(int i=0;i<que[u].size();i++)
    {
        int v = que[u][i].v;
        if(vis[v])
        {
            num[que[u][i].c]=dis[v]+dis[u]-2*dis[Find(v)];
        }
    }
    for(int i=0;i<tree[u].size();i++)
    {
        int v=tree[u][i].v;
        if(!vis[v])
        {
            dis[v] = dis[u]+tree[u][i].c;
            lca(v);
            f[v] = u;
        }
    }
}

int main()
{
    int x,y,c,n,m,q;
    char s[5];
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        Init(n);
        for(int i=0;i<m;i++)
        {
            scanf("%d %d %d %s",&x,&y,&c,s);
            node temp;
            temp.v = y;
            temp.c = c;
            tree[x].push_back(temp);
            temp.v = x;
            tree[y].push_back(temp);
        }
        scanf("%d",&q);
        for(int i=0;i<q;i++)
        {
            scanf("%d %d",&x,&y);
            node temp;
            temp.v = y;
            temp.c = i;
            que[x].push_back(temp);
            temp.v = x;
            que[y].push_back(temp);
        }
        lca(1);
        for(int i=0;i<q;i++)
            printf("%d\n",num[i]);
    }

}
时间: 08-20

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