Total Highway Distance

Total Highway Distance

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

Little Hi and Little Ho are playing a construction simulation game. They build N cities (numbered from 1 to N) in the game and connect them by N-1 highways. It is guaranteed that each pair of cities are connected by the highways directly or indirectly.

The game has a very important value called Total Highway Distance (THD) which is the total distances of all pairs of cities. Suppose there are 3 cities and 2 highways. The highway between City 1 and City 2 is 200 miles and the highway between City 2 and City 3 is 300 miles. So the THD is 1000(200 + 500 + 300) miles because the distances between City 1 and City 2, City 1 and City 3, City 2 and City 3 are 200 miles, 500 miles and 300 miles respectively.

During the game Little Hi and Little Ho may change the length of some highways. They want to know the latest THD. Can you help them?

输入

Line 1: two integers N and M.

Line 2 .. N: three integers u, v, k indicating there is a highway of k miles between city u and city v.

Line N+1 .. N+M: each line describes an operation, either changing the length of a highway or querying the current THD. It is in one of the following format.

EDIT i j k, indicating change the length of the highway between city i and city j to k miles.

QUERY, for querying the THD.

For 30% of the data: 2<=N<=100, 1<=M<=20

For 60% of the data: 2<=N<=2000, 1<=M<=20

For 100% of the data: 2<=N<=100,000, 1<=M<=50,000, 1 <= u, v <= N, 0 <= k <= 1000.

输出

For each QUERY operation output one line containing the corresponding THD.

样例输入

3 5
1 2 2
2 3 3
QUERY
EDIT 1 2 4
QUERY
EDIT 2 3 2
QUERY

样例输出

10
14
12

分析：每条边对答案的贡献都是两边点个数的乘积乘上这个长度，然后维护好长度来更新答案；

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=1e5+10;
const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}};
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int n,m,k,t,val[maxn],son[maxn],fa[maxn];
ll ans;
vi a[maxn],b[maxn];
char q[10];
int dfs(int now,int pre)
{
    son[now]=1;
    for(int i=0;i<a[now].size();i++)
    {
        int x=a[now][i],y=b[now][i];
        if(x!=pre)
        {
            son[now]+=dfs(x,now);
            fa[x]=now,val[x]=y;
            ans+=(ll)son[x]*(n-son[x])*val[x];
        }
    }
    return son[now];
}
int main()
{
    int i,j;
    scanf("%d%d",&n,&m);
    rep(i,1,n-1)
    {
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        a[u].pb(v),a[v].pb(u);
        b[u].pb(w),b[v].pb(w);
    }
    dfs(1,-1);
    rep(i,1,m)
    {
        scanf("%s",q);
        if(q[0]=='Q')printf("%lld
",ans);
        else
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            if(fa[v]==u)swap(u,v);
            ans=ans+(ll)son[u]*(n-son[u])*(w-val[u]);
            val[u]=w;
        }
    }
    //system("Pause");
    return 0;
}