[APIO 2010] 巡逻

[题目链接]

https://www.lydsy.com/JudgeOnline/problem.php?id=1912

[算法]

树的直径

[代码]

#include<bits/stdc++.h>
using namespace std;
#define MAXN 100010

int i,n,l1,l2,tot,u,v,k,now;
int head[MAXN],pre[MAXN],d[MAXN];
vector< int > path;
pair< int,pair<int,int> > tmp;

struct edge
{
        int to,w,nxt;
} e[MAXN << ];

inline void addedge(int u,int v,int w)
{
        tot++;
        e[tot] = (edge){v,w,head[u]};
        head[u] = tot;
}
inline void dfs1(int u,int fa)
{
        int i,v,w;
        for (i = head[u]; i; i = e[i].nxt)
        {
                v = e[i].to;
                w = e[i].w;
                if (v == fa) continue;
                dfs1(v,u);
                if (d[u] + d[v] + w  > l1)
                {
                        l1 = d[u] + d[v] + w;
                        tmp = make_pair(i,make_pair(pre[u],pre[v]));
                }
                if (d[v] + w > d[u])
                {
                        d[u] = d[v] + w;
                        pre[u] = i;
                }
        }
}
inline void dfs2(int u,int fa)
{
        int i,v,w;
        for (i = head[u]; i; i = e[i].nxt)
        {
                v = e[i].to;
                w = e[i].w;
                if (v == fa) continue;
                dfs2(v,u);
                if (d[u] + d[v] + w > l2) l2 = d[u] + d[v] + w;
                d[u] = max(d[u],d[v] + w);
        }
}
int main()
{

        scanf("%d%d",&n,&k);
        for (i = ; i < n; i++)
        {
                scanf("%d%d",&u,&v);
                addedge(u,v,);
                addedge(v,u,);
        }
        l1 = ;
        memset(d,,sizeof(d));
        dfs1(,);
        now = tmp.second.first;
        while (now != )
        {
                path.push_back(now);
                now = pre[e[now].to];
        }
        path.push_back(tmp.first);
        now = tmp.second.second;
        while (now != )
        {
                path.push_back(now);
                now = pre[e[now].to];
        }
        for (i = ; i < path.size(); i++)
        {
                e[path[i]].w *= -;
                e[path[i] ^ ].w *= -;
        }
        l2 = ;
        memset(d,,sizeof(d));
        dfs2(,);
        if (k == ) printf("%d\n", * (n - ) - l1 + );
        else printf("%d\n", * n - l1 - l2);

        return ;

}