<span>hihocoder1238 Total Highway Distance(树形dp)</span>
题意:
给定一颗有N个节点的带权树,之后进行M次操作:
Q操作:询问树上所有点对之间的距离之和
E操作:修改树上某一条边的权值
思路:
树形dp求出每条边被利用的次数并统计
然后修改的时候就把这个边权修改并将改变值乘上利用次数更改ans
/* *********************************************** Author :devil ************************************************ */ #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <cmath> #include <stdlib.h> #define inf 0x3f3f3f3f #define LL long long #define rep(a,b) for(int i=a;i<b;i++) #define dec(a,b) for(int i=a;i>=b;i--) #define ou(a) printf("%d\n",a) #define pb push_back #define mkp make_pair template<class T>inline void rd(T &x){char c=getchar();x=0;while(!isdigit(c))c=getchar();while(isdigit(c)){x=x*10+c-'0';c=getchar();}} #define IN freopen("in.txt","r",stdin); #define OUT freopen("out.txt","w",stdout); using namespace std; const int mod=1e9+7; const int N=1e5+10; int n,m,x,y,w,sum[N],val[N],level[N]; LL ans; char s[10]; vector<pair<int,int> >eg[N]; void dfs(int u,int lev,int pre) { sum[u]=1; level[u]=lev; rep(0,eg[u].size()) { int v=eg[u][i].first; if(v==pre) continue; val[v]=eg[u][i].second; dfs(v,lev+1,u); sum[u]+=sum[v]; ans+=(LL)val[v]*sum[v]*(n-sum[v]); } } int main() { //IN rd(n);rd(m); rep(1,n) { rd(x),rd(y),rd(w); eg[x].pb(mkp(y,w)); eg[y].pb(mkp(x,w)); } dfs(1,1,1); while(m--) { scanf("%s",s); if(s[0]=='Q') printf("%lld\n",ans); else { rd(x),rd(y),rd(w); if(level[x]<level[y]) swap(x,y); ans+=(LL)(w-val[x])*sum[x]*(n-sum[x]); val[x]=w; } } return 0; }