POJ 1987 Distance Statistics 树分治

最新推荐文章于 2021-01-22 12:15:41 发布

转载最新推荐文章于 2021-01-22 12:15:41 发布 · 140 阅读

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原文链接：http://www.cnblogs.com/zxhl/p/5692947.html

本文针对距离统计问题，提出了一种有效的算法解决方案。该方案通过构建图结构并运用DFS深度优先搜索算法来计算任意两个农场之间的最短路径，最终统计出所有距离不超过K的农场对数目。

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Distance Statistics

Description

Frustrated at the number of distance queries required to find a reasonable route for his cow marathon, FJ decides to ask queries from which he can learn more information. Specifically, he supplies an integer K (1 <= K <= 1,000,000,000) and wants to know how many pairs of farms lie at a distance at most K from each other (distance is measured in terms of the length of road required to travel from one farm to another). Please only count pairs of distinct farms (i.e. do not count pairs such as (farm #5, farm #5) in your answer).

Input

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

* Line M+2: A single integer, K.

Output

* Line 1: The number of pairs of farms that are at a distance of at most K from each-other.

Sample Input

Sample Output

Hint

There are 5 roads with length smaller or equal than 10, namely 1-4 (3), 4-7 (2), 1-7 (5), 3-5 (7) and 3-6 (9).

题解：

　　POJ 1741

　　http://www.cnblogs.com/zxhl/p/5692688.html

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
const int N = 4e4+20, M = 1e2+10, mod = 1e9+7, inf = 1e9+1000;
typedef long long ll;

int ans, n,m,root , t = 1,K,siz[N],head[N],f[N],deep[N],d[N],allnode,vis[N];
struct edg{int to,next,v,w;}e[N * 4];
void add(int u,int v,int w) {e[t].to=v;e[t].v=w;e[t].next=head[u];head[u]=t++;}

void getroot(int x,int fa) {
    siz[x] = 1;
    f[x] = 0;
    for(int i=head[x];i;i=e[i].next) {
        int to = e[i].to;
        if(to == fa || vis[to]) continue;
        getroot(to,x);
        siz[x] += siz[to];
        f[x] = max(f[x] , siz[to]);
    }
    f[x] = max(f[x] , allnode - siz[x]);
    if(f[x] < f[root]) root = x;
}
void getdeep(int x,int fa) {
    if(d[x] <= K) deep[++deep[0]]=d[x];
    for(int i=head[x];i;i=e[i].next) {
        int to = e[i].to;
        if(to == fa || vis[to]) continue;
        d[to] = d[x] + e[i].v;
        getdeep(to,x);
    }
}
int cal(int x,int now) {
    d[x]=now;deep[0] = 0;
    getdeep(x,0);
    sort(deep+1,deep+deep[0]+1);
    int all = 0;
    for(int l=1,r=deep[0];l<r;) {
        if(deep[l]+deep[r] <= K) {all+=r-l;l++;}
        else r--;
    }
    return all;
}
void work(int x) {
    ans+=cal(x,0);
    vis[x] = 1;
    for(int i=head[x];i;i=e[i].next) {
        int to = e[i].to;
        if(vis[to]) continue;
        ans-=cal(to,e[i].v);
        allnode = siz[to];
        root = 0;
        getroot(to,root);
        work(root);
    }
}
void init()
{
    memset(head,0,sizeof(head));
    t = 1;
    ans = root = 0;
    memset(vis,0,sizeof(vis));
}
int main()
{
    while(~scanf("%d%d",&n,&m)) {
        init();
        for(int i=1;i<n;i++) {
            int a,b,c;char ch[2];
            scanf("%d%d%d%s",&a,&b,&c,ch);
            add(a,b,c) , add(b,a,c);
        }
        scanf("%d",&K);
        allnode=n;f[0]=inf;
        getroot(1,0);
        work(root);
        printf("%d\n",ans);
    }

}