1663: TreeTime Limit: 5 SecMemory Limit: 128 MBSubmit: 26Solved: 11[Submit][Status][Web Board]Description
CSU has a lot of trees. But there is a tree which is different from the others. This one is made of weighted edges and I have three kinds of operations on it:1. C a b: Change the weight of edge a to b (1 ≤ b ≤ 100);2. M a b c: Multiply the weights of those edges on the path from node a to node b by c (|c|≤10, c ≠ 0);3. Q a b: Get the sum of weights from all the edges on the path from node a to node b.
Input
There are multiple test cases.The first line will contain a positive integer T (T ≤ 10) meaning the number of test cases.Each test case will have an integer N (1 ≤ N ≤ 50,000) indicating the number of nodes marked from 1 to N.Then N-1 lines followed. Each lines contains three integers a, b and c (1 ≤ a, b, c ≤ N, a ≠ b, 1 ≤ c ≤ 100) indicating the there is an edge connecting node a and node b with weight c. The edges are marked from 1 to N-1 in the order of appearance.Then some operations followed, with one operation per line. The format is as shown in the problem description. A single letter ‘E’ indicates the end of the case. There are no more than 50,000 operations per test case.
Output
For each ‘Q’ operation, output an integer meaning the sum. The final result will never exceed 32-bit signed integer.
Sample Input241 2 141 3 203 4 1C 3 11M 1 4 3M 3 4 2M 1 3 2Q 2 4M 1 4 -1M 3 4 -2Q 2 4E31 2 41 3 2Q 2 3C 1 7M 2 3 2M 1 3 5Q 2 3ESample Output20026634#include<stdio.h> #include<string.h> #define LL long long const int N = 50005;int head[N<<1],to[N<<1],next1[N<<1],tot; int deep[N],fath[N],son[N],num[N]; int top[N],p[N],pos;void init(){pos=tot=0;memset(head,-1,sizeof(head)); } void addEdge(const int& u, const int& v){to[tot] = v, next1[tot] = head[u], head[u] = tot++; } void addUndirEdge(const int& u, const int& v){addEdge(u, v), addEdge(v, u); }void dfs1(int u,int pre,int d){fath[u]=pre;deep[u]=d;son[u]=-1;num[u]=1;for(int i=head[u]; i!=-1; i=next1[i]){int v=to[i];if(v==fath[u])continue;dfs1(v,u,d+1);num[u]+=num[v];if(son[u]==-1||num[v]>num[son[u]])son[u]=v;} } void getpos(int u,int root){top[u]=root;p[u]=pos++;if(son[u]==-1)return ;getpos(son[u],root);for(int i=head[u]; i!=-1; i=next1[i]){int v=to[i];if(v==son[u]||v==fath[u])continue;getpos(v,v);} } struct TREE{LL milt,sum; }root[N*3]; LL cost[N];void pushUp(int k){root[k].sum=root[k<<1].sum+root[k<<1|1].sum; } void pushDow(int k){if(root[k].milt!=1){root[k<<1].sum*=root[k].milt;root[k<<1].milt*=root[k].milt;root[k<<1|1].sum*=root[k].milt;root[k<<1|1].milt*=root[k].milt;root[k].milt=1;} } void build(int l, int r, int k){root[k].milt=1;if(l==r){root[k].sum=cost[l]; return ;}int mid=(l+r)>>1;build(l,mid,k<<1);build(mid+1,r,k<<1|1);pushUp(k); } void update_C(int l, int r, int k, const int& id, LL c){if(l==r){root[k].sum=c; return ;}int mid=(l+r)>>1;pushDow(k);if(id<=mid)update_C(l,mid,k<<1,id,c);elseupdate_C(mid+1,r,k<<1|1,id,c);pushUp(k); } void updata_M(int l,int r,int k,int L,int R,int M){if(L<=l&&r<=R){root[k].milt*=M; root[k].sum*=M;return ;}pushDow(k);int mid=(l+r)>>1;if(L<=mid)updata_M(l,mid,k<<1,L,R,M);if(mid<R)updata_M(mid+1,r,k<<1|1,L,R,M);pushUp(k); } LL query(int l, int r, int k, const int& L, const int& R){if(L<=l&&r<=R){return root[k].sum;}pushDow(k);int mid=(l+r)>>1;LL sum=0;if(L<=mid)sum+=query(l,mid,k<<1,L,R);if(mid<R)sum+=query(mid+1,r,k<<1|1,L,R);return sum; } void swp(int &u,int &v){int tt=u; u=v; v=tt; } LL solve(int u,int v,int flag,LL M){int fu=top[u], fv=top[v];LL sum=0;while(fu!=fv){if(deep[fu]<deep[fv]){swp(fu,fv); swp(u,v);}if(flag==0)updata_M(1,pos,1,p[fu],p[u],M);elsesum+=query(1,pos,1,p[fu],p[u]);u=fath[fu]; fu=top[u];}if(u==v)return sum;if(deep[u]>deep[v])swp(u,v);if(flag==0)updata_M(1,pos,1,p[son[u]],p[v],M);elsesum+=query(1,pos,1,p[son[u]],p[v]);//一不小心p[son[u]]写成了p[u]让我WA了好几次（求边权用p[son[u]],求点权用p[u]）return sum; }struct EDG{int u,v;LL c; }edg[N];int main() {int n,m,a,b,T;char op[10];scanf("%d",&T);while(T–){scanf("%d",&n);init();for(int i=1; i<n; i++){scanf("%d%d%lld",&edg[i].u,&edg[i].v,&edg[i].c);addUndirEdge(edg[i].u, edg[i].v);}dfs1(1,1,1);getpos(1,1);for(int i=1; i<n; i++){if(deep[edg[i].u]>deep[edg[i].v])swp(edg[i].u, edg[i].v);cost[p[edg[i].v]]=edg[i].c;}pos=n;build(1,pos,1);while(1){scanf("%s",op);if(op[0]=='E')break;scanf("%d%d",&a,&b);if(op[0]=='C')update_C(1,pos,1,p[edg[a].v],b);else if(op[0]=='M'){LL M;scanf("%lld",&M);solve(a,b,0,M);}elseprintf("%lld\n",solve(a,b,1,1));}}}/**************************************************************Problem: 1663User: aking2015Language: C++Result: AcceptedTime:2696 msMemory:9044 kb ****************************************************************/
，接受失败等于打破完美的面具，接受失败等于放松自己高压的心理，
