luogu1131 [ZJOI2007]时态同步

Posted 2020-10-20 poorpool

num[x]表示x到达叶子最远路径。
每个子节点对答案的贡献是num[x] - (num[t] + edge[i].val)

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long ll;
int n, ss, hea[500005], cnt, uu, vv, ww;
ll num[500005], ans;
struct Edge{
    int too, nxt, val;
}edge[1000005];
void add_edge(int fro, int too, int val){
    edge[++cnt].nxt = hea[fro];
    edge[cnt].too = too;
    edge[cnt].val = val;
    hea[fro] = cnt;
}
void dfs(int x, int f){
    for(int i=hea[x]; i; i=edge[i].nxt){
        int t=edge[i].too;
        if(t!=f){
            dfs(t, x);
            num[x] = max(num[x], num[t]+edge[i].val);
        }
    }
    for(int i=hea[x]; i; i=edge[i].nxt){
        int t=edge[i].too;
        if(t!=f)
            ans += num[x] - (num[t] + edge[i].val);
    }
}
int main(){
    cin>>n>>ss;
    for(int i=1; i<n; i++){
        scanf("%d %d %d", &uu, &vv, &ww);
        add_edge(uu, vv, ww);
        add_edge(vv, uu, ww);
    }
    dfs(ss, 0);
    cout<<ans<<endl;
    return 0;
}