P1131 [ZJOI2007]时态同步

${color{cyan}{>>Question}}$

题意让我们用最少的代价把叶子节点到根节点的距离调成相同

每次向最长链看齐即可

令$f[u]$表示以$u$为根到叶节点的最长链长度

则$$f[u] = maxleft{f[v]+e(u,v) ight}$$

统计答案

$$ans = sum _{v(e(u,v)in E)} f[u]-(f[v]+e(u,v))$$

上代码

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#define ll long long
using namespace std; 

template <typename T> void in(T &x) {
    x = 0; T f = 1; char ch = getchar();
    while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
    while( isdigit(ch)) {x = 10 * x + ch - 48; ch = getchar();}
    x *= f;
}

template <typename T> void out(T x) {
    if(x < 0) x = -x , putchar('-');
    if(x > 9) out(x/10);
    putchar(x%10 + 48);
}
//-------------------------------------------------------

const int N = 5e5+7;

int n,s;
ll ans,f[N];

struct edge {
    int v,w,nxt;
    edge(int v = 0,int w = 0,int nxt = 0):v(v),w(w),nxt(nxt){};
}e[N<<1]; int head[N],e_cnt;

void add(int u,int v,int w) {
    e[++e_cnt] = edge(v,w,head[u]); head[u] = e_cnt;
}

void dfs(int u,int fa) {
    int i;
    int max_e = 0;
    for(i = head[u]; i;i = e[i].nxt) {
        int v = e[i].v; if(v == fa) continue;
        dfs(v,u);
        f[u] = max(f[u],f[v]+e[i].w);
    }
    for(i = head[u]; i;i = e[i].nxt) {
        int v = e[i].v; if(v == fa) continue;
        ans += f[u] - (f[v]+e[i].w);
    }
}

int main() {
    int i,u,v,w;
    in(n); in(s);
    for(i = 1;i < n; ++i) {
        in(u); in(v); in(w);
        add(u,v,w); add(v,u,w);
    }
    dfs(s,0);
    out(ans);
    return 0;
}