zoj 3820 Building Fire Stations(贪心 + 二分)

最新推荐文章于 2021-08-18 21:36:26 发布

原创最新推荐文章于 2021-08-18 21:36:26 发布 · 918 阅读

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该博客介绍了一个树形结构问题，目标是找到两个点以使树上所有点到这两点之一的最短路径最大值最小。解决策略涉及贪心思想和二分查找技术。首先，确定一点作为中心，然后二分搜索最短路径的最大值，通过BFS验证找到的点是否满足条件。

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题意：在树上找两个点，使得树上的点到这两个点之一的最短距离的最大值最小。

思路：一个点的话明显就是中心了。换成了两个点，那么先考虑一下，明显这两个点在直径上是最优的，那么接下来，就可以二分一下距离，这样就可以求出相应的点的位置，接下来通过bfs判断是否满足要求就行了。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<set>
#include<cmath>
#include<vector>
#include<bitset>
#define inf 0x3f3f3f3f
#define Inf 0x3FFFFFFFFFFFFFFFLL
#define eps 1e-6
#define pi acos(-1.0)
using namespace std;
typedef long long ll;
const int maxn = 200000 + 10;
struct Edge
{
    int v,next;
    Edge(int v = 0,int next = 0):v(v),next(next){}
}edges[maxn<<1];
int head[maxn],d[maxn],pa[maxn],S[maxn],nEdge;
int rp[maxn],vis[maxn],cnt,ux,uy,ans,m;
void AddEdges(int u,int v)
{
    edges[++nEdge] = Edge(v,head[u]);
    head[u] = nEdge;
    edges[++nEdge] = Edge(u,head[v]);
    head[v] = nEdge;
}
bool check(int limit)
{
    cnt++;
    int a = rp[limit + 1];
    int b = rp[m - limit];
    if(limit + 1 >= m - limit)
    {
        a = rp[m/2];
        b = rp[m/2 + 1];
    }
    int tail = 0,u,v;
    d[a] = d[b] = 0;
    S[tail++] = a;
    S[tail++] = b;
    vis[a] = vis[b] = cnt;
    for(int i = 0;i < tail;++i)
    {
        u = S[i];
        if(d[u] > limit) return false;
        for(int k = head[u];k != -1;k = edges[k].next)
        {
            v = edges[k].v;
            if(vis[v] == cnt) continue;
            vis[v] = cnt;
            pa[v] = u;
            d[v] = d[u] + 1;
            S[tail++] = v;
        }
    }
    if(limit < ans)
    {
        ans = limit;
        ux = a;
        uy = b;
    }
    return true;
}
void solve()
{
    pa[1] = -1;
    int tail = 0,maxd = 0,mpos,u,v;
    S[tail++] = 1;
    d[1] = 0;
    for(int i = 0;i < tail;++i)
    {
        u = S[i];
        if(d[u] > maxd)
        {
            maxd = d[u];
            mpos = u;
        }
        for(int k = head[u];k != -1;k = edges[k].next)
        {
            v = edges[k].v;
            if(v == pa[u]) continue;
            d[v] = d[u] + 1;
            pa[v] = u;
            S[tail++] = v;
        }
    }
    pa[mpos] = -1;
    tail = 0;
    S[tail++] = mpos;
    d[mpos] = 0;
    maxd = 0;
    for(int i = 0;i < tail;++i)
    {
        u = S[i];
        if(d[u] > maxd)
        {
            maxd = d[u];
            mpos = u;
        }
        for(int k = head[u];k != -1;k = edges[k].next)
        {
            v = edges[k].v;
            if(v == pa[u]) continue;
            pa[v] = u;
            d[v] = d[u] + 1;
            S[tail++] = v;
        }
    }
    m = 0;
    u = mpos;
    while(u != -1)
    {
        rp[++m] = u;
        u = pa[u];
    }
    int L = 0,R = tail,mid;
    ans = inf;
    cnt = 0;
    memset(vis,0,sizeof(vis));
    while(L < R)
    {
        mid = (L + R)>>1;
        if(check(mid)) R = mid;
        else L = mid + 1;
    }
    check(L);
}
int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(head,0xff,sizeof(head));
        nEdge = -1;
        int n,u,v;
        scanf("%d",&n);
        for(int i = 1;i < n;++i)
        {
            scanf("%d%d",&u,&v);
            AddEdges(u,v);
        }
        solve();
        printf("%d %d %d\n",ans,ux,uy);
    }
    return 0;
}