题目链接:https://www.luogu.org/problemnew/show/P2491
题外话:
OI一共只有三种题——会的题,不会的题,二分题。
题解:
step 1 求树的直径,把树的直径上的路径边权都置为0,这样了再求一次其他点最短路。
step 2 在树的直径上二分,具体方法是把树的直径长度用类似前缀和的思想处理后,二分左右端点舍去的距离。
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 300010;
int n, dis[maxn], ans = 0x7fffffff, node, pre[maxn], path[maxn], tot, D, sta[maxn], top, s, S;
bool vis[maxn];
struct edge{
int from, to, next, len;
}e[maxn<<2];
int cnt, head[maxn];
void add(int u, int v, int w)
{
e[++cnt].from = u;
e[cnt].next = head[u];
e[cnt].len = w;
e[cnt].to = v;
head[u] = cnt;
}
queue<int> q;
void bfs(int s)
{
vis[s] = 1;
q.push(s);
while(!q.empty())
{
int now = q.front(); q.pop();
for(int i = head[now]; i != -1; i = e[i].next)
{
if(vis[e[i].to] == 0)
{
pre[e[i].to] = now;
dis[e[i].to] = dis[now] + e[i].len;
vis[e[i].to] = 1;
q.push(e[i].to);
if(dis[e[i].to] > D)
{
D = dis[e[i].to];
node = e[i].to;
}
}
}
}
}
void SPFA(int s)
{
memset(vis, 0, sizeof(vis));
memset(dis, 127, sizeof(dis));
dis[s] = 0;
vis[s] = 1;
q.push(s);
while(!q.empty())
{
int now = q.front(); q.pop();
vis[now] = 0;
for(int i = head[now]; i != -1; i = e[i].next)
{
if(dis[e[i].to] > dis[now] + e[i].len)
{
dis[e[i].to] = dis[now] + e[i].len;
if(!vis[e[i].to])
{
q.push(e[i].to);
vis[e[i].to] = 1;
}
}
}
}
}
bool check(int D)
{
int l = 1, r = top;
while(sta[1] - sta[l+1] <= D && l <= top) l++;
while(sta[r-1] <= D && r >= 1) r--;
//cout<<l<<" "<<r<<endl;
return sta[l] - sta[r] <= S;
}
int main()
{
memset(head, -1, sizeof(head));
int u, v, w;
cin>>n>>S;
for(int i = 1; i < n; i++)
{
scanf("%d%d%d",&u, &v, &w);
add(u, v, w);
add(v, u, w);
}
memset(dis, 0, sizeof(dis));
memset(vis, 0, sizeof(vis));
D = 0;
bfs(1);
int s = node;
memset(vis, 0, sizeof(vis));
dis[node] = 0;
D = 0;
bfs(s);
int x = node;
D = dis[x];
sta[++top] = D;
path[++tot] = x;
while(x != s)
{
path[++tot] = pre[x];
sta[++top] = dis[pre[x]];
x = pre[x];
}
//for(int i = 1; i <= n; i++) cout<<path[i]<<" ";
for(int i = 2; i <= tot; i++)
{
add(path[i-1], path[i], 0);
add(path[i], path[i-1], 0);
}
SPFA(x);
int l = 0, r = D;
for(int i = 1; i <= n; i++) l = max(l, dis[i]);
if(S < D)
{
while(l <= r)
{
int mid = (l + r) >> 1;
if(check(mid)) r = mid - 1;
else l = mid + 1;
}
}
//for(int i = 1; i <= n; i++) cout<<dis[i];
printf("%d\n", l);
return 0;
}
附:
sdoi居然出提高组原题(#滑稽 不过是数据扩大了1000倍
据说那个题想怎么暴怎么暴(逃