bzoj3124: [Sdoi2013]直径 树形dp two points

题目链接

bzoj3124: [Sdoi2013]直径

题解

发现所有直径都经过的边

一定在一条直径上，并且是连续的

在一条直径上找这段区间的两个就好了

代码

#include<map>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define gc getchar()
#define pc putchar
#define int long long
inline int read() {
    int x = 0,f = 1;
    char c = gc;
    while(c < '0' || c > '9') { if(c == '-') f =- 1;c = gc;}
    while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = gc;
    return x * f;
}
void print(int x) {
    if(x < 0) {
        pc('-');
        x = -x;
    }
    if(x >= 10) print(x / 10);
    pc(x % 10 + '0');
}
int n;
const int maxn = 400007;
struct node {
    int v,w,next;
} edge[maxn << 1];
int head[maxn],num = 0;
inline void add_edge(int u,int v,int w) {
    edge[++ num].v = v; edge[num].w = w; edge[num].next = head[u]; head[u] = num;
}
int st,mx,dis[maxn],fa[maxn];
void dfs(int x,int Fa) {
    if(dis[x] > mx) {
        mx = dis[x]; st = x;
    }
    fa[x] = Fa;
    for(int i = head[x];i;i = edge[i].next) {
        int v = edge[i].v;
        if(v == Fa) continue;
        dis[v] = dis[x] + edge[i].w;
        dfs(v,x);
    }
}
int chain[maxn];
int rdis[maxn],MX;
bool vis[maxn];
void rdfs(int x,int Fa) {
    vis[x] = true;
    MX = std::max(MX,rdis[x]);
    for(int i = head[x];i;i = edge[i].next) {
        int v = edge[i].v;
        if(vis[v] || v == Fa) continue;
        rdis[v] = rdis[x] + edge[i].w;
        rdfs(v,x);
    }
}
main() {
    //freopen("3.in","r",stdin);
     n = read();
    for(int u,v,w,i = 1;i < n;++ i) {
        u = read(),v = read(),w = read();
        add_edge(u,v,w);
        add_edge(v,u,w);
    }
    dfs(1,0);
    dis[st] = 0;
    mx = 0;
    dfs(st,0);
    int len = 0;
    while(st) {
        chain[++ len] = st;
        vis[st] = true;
        st = fa[st];
    }
    int l = len,r = 1;
    for(int i = len;i >= 1;-- i) {
        MX = 0;
        rdfs(chain[i],0);
        if(!MX) continue;
        if(MX == dis[chain[i]]) l = i;
        if(MX == mx - dis[chain[i]]) {r = i;break;}
    }
    print(mx);
    pc('\n');
    print(l - r);
}
/*
6
3 1 1000
1 4 10
4 2 100
4 5 50
4 6 100
*/