bzoj2282

 

树的直径+单调队列

这竟然和bzoj1999是一样的？

我们yy一下，发现这条路径肯定在树的直径上，然后就好办了，我们维护一个双指针，保证长度<=s,然后最大距离就是直径的两端到路径的两端的最大值，还有当前路径上挂着的链，这个我们dfs一下就行了，然后直径两端的最大值直接求就行了，链的最大值维护一个单调队列，那么就可以O(n)出解了，求直径用两次bfs

bzoj1999卡到rank1了！

bzoj2282rank2！

#include<bits/stdc++.h>
using namespace std;
const int N = 500010;
struct edge {
    int nxt, to, w;
} e[N << 1];
int n, cnt = 1, S, ans = 0x3f3f3f3f;
int d[N], route[N], head[N], vis[N], dis[N], dep[N], q[N], last[N];
namespace IO 
{
    const int Maxlen = N * 50;
    char buf[Maxlen], *C = buf;
    int Len;
    inline void read_in()
    {
        Len = fread(C, 1, Maxlen, stdin);
        buf[Len] = '\0';
    }
    inline void fread(int &x) 
    {
        x = 0;
        int f = 1;
        while (*C < '0' || '9' < *C) { if(*C == '-') f = -1; ++C; }
        while ('0' <= *C && *C <= '9') x = (x << 1) + (x << 3) + *C - '0', ++C;
        x *= f;
    }
    inline void read(int &x)
    {
        x = 0;
        int f = 1; char c = getchar();
        while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
        while(c >= '0' && c <= '9') { x = (x << 1) + (x << 3) + c - '0'; c = getchar(); }
        x *= f;
    }
    inline void read(long long &x)
    {
        x = 0;
        long long f = 1; char c = getchar();
        while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
        while(c >= '0' && c <= '9') { x = (x << 1ll) + (x << 3ll) + c - '0'; c = getchar(); }
        x *= f;
    } 
} using namespace IO;
void link(int u, int v, int w)
{
    e[++cnt].nxt = head[u];
    head[u] = cnt;
    e[cnt].to = v;
    e[cnt].w = w;
}
int bfs(int s)
{
    int ret = s, l = 1, r = 0;
    d[s] = 0;
    q[++r] = s;
    last[s] = 0;
    while(l <= r)
    {
        int u = q[l++];
        for(int i = head[u]; i; i = e[i].nxt) if(e[i].to != e[last[u] ^ 1].to)
        {
            d[e[i].to] = d[u] + e[i].w;
            if(d[e[i].to] > d[ret]) ret = e[i].to;
            q[++r] = e[i].to;
            last[e[i].to] = i;
        }
    }
    return ret;
}
int dfs(int u)
{
    int ret = 0;
    vis[u] = 1;
    for(int i = head[u]; i; i = e[i].nxt) if(!vis[e[i].to]) ret = max(ret, dfs(e[i].to) + e[i].w);
    return ret; 
}
int main()
{
    read_in();
    fread(n);
    fread(S);
    for(int i = 1; i < n; ++i)
    {
        int u, v, w;
        fread(u);
        fread(v);
        fread(w);
        link(u, v, w);
        link(v, u, w);
    }
    int s = bfs(1), t = bfs(s), now = t, p = 0, l = 1, r = 0;
    while(now != s) route[++route[0]] = now, vis[now] = 1, now = e[last[now] ^ 1].to;
    route[++route[0]] = s;
    vis[s] = 1;
    for(int i = 1; i <= route[0]; ++i) 
    {
        int tmp = i == 1 ? 0 : e[last[route[i - 1]]].w;
        dis[i] = dis[i - 1] + tmp;
        dep[i] = dfs(route[i]);
    }   
    for(int i = 1; i <= route[0]; ++i)
    {
        int mx = dis[i];
        while(l <= r && q[l] < i) ++l;
        while(dis[p + 1] - dis[i] <= S && p < route[0]) 
        {
            ++p;    
            while(l <= r && dep[p] > dep[q[r]]) --r;
            q[++r] = p;
        }
        mx = max(mx, max(dis[route[0]] - dis[p], dep[q[l]]));
        ans = min(ans, mx); 
    }
    printf("%d\n", ans);
    return 0;
}