树链剖分+离散+扫描（HDU5044）

Tree

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1385    Accepted Submission(s): 237

Problem Description

You are given a tree (an acyclic undirected connected graph) with N nodes. The tree nodes are numbered from 1 to N

There are N - 1 edges numbered from 1 to N - 1.

Each node has a value and each edge has a value. The initial value is 0.

There are two kind of operation as follows:

● ADD1 u v k: for nodes on the path from u to v, the value of these nodes increase by k.

● ADD2 u v k: for edges on the path from u to v, the value of these edges increase by k.

After finished M operation on the tree, please output the value of each node and edge.

 

Input

The first line of the input is T (1 ≤ T ≤ 20), which stands for the number of test cases you need to solve.

The first line of each case contains two integers N ,M (1 ≤ N, M ≤10⁵),denoting the number of nodes and operations, respectively.

The next N - 1 lines, each lines contains two integers u, v(1 ≤ u, v ≤ N ), denote there is an edge between u,v and its initial value is 0.

For the next M line, contain instructions “ADD1 u v k” or “ADD2 u v k”. (1 ≤ u, v ≤ N, -10⁵ ≤ k ≤ 10⁵)

 

Output

For each test case, print a line “Case #t:”(without quotes, t means the index of the test case) at the beginning.

The second line contains N integer which means the value of each node.

The third line contains N - 1 integer which means the value of each edge according to the input order.

 

Sample Input

24 21 22 32 4ADD1 1 4 1ADD2 3 4 24 21 22 31 4ADD1 1 4 5ADD2 3 2 4

 

Sample Output

Case #1:1 1 0 10 2 2Case #2:5 0 0 50 4 0
题意：给出一棵树，树上的点和边的权值开始都是0，有两种操作，对于第一种操作，ADD1：在u到v的路径上每个点的权值+w；对于第二种操作ADD2：在u到v的路径上每个边的权值+w；最后询问每个点的权值和每条边的权值（按照输入的顺序）
分析：首先两个dfs进行轻重链剖分，然后对于每个区间[L,R],在L的位置+w，在R+1的位置-w，保存在g1数组中，同理，对于边也一样，保存在g2数组里，最后从前往后做一遍扫描即可；
程序：
#pragma comment(linker, "/STACK:1024000000,1024000000")#include"stdio.h"#include"string.h"#include"iostream"#include"map"#include"string"#include"queue"#include"stdlib.h"#include"algorithm"#include"math.h"#define M 110009#define eps 1e-5#define inf 100000000#define mod 100000000#define INF 0x3f3f3f3fusing namespace std;struct node{    int v;    node(int vv){v=vv;}};vector<node>edge[M];int pos,son[M],fa[M],p[M],fp[M],deep[M],top[M],num[M],g1[M],g2[M],ans[M];void dfs(int u,int f,int d){    deep[u]=d;    fa[u]=f;    num[1]=1;    for(int i=0;i<(int)edge[u].size();i++)    {        int v=edge[u][i].v;        if(v==f)continue;        dfs(v,u,d+1);        num[u]+=num[v];        if(son[u]==-1||num[son[u]]<son[v])            son[u]=v;    }}void getpos(int u,int sp){    top[u]=sp;    p[u]=pos++;    fp[p[u]]=u;    if(son[u]==-1)return;    getpos(son[u],sp);    for(int i=0;i<(int)edge[u].size();i++)    {        int v=edge[u][i].v;        if(v==son[u]||v==fa[u])continue;        getpos(v,v);    }}void init(){    pos=0;    memset(son,-1,sizeof(son));    dfs(1,1,1);    getpos(1,1);}void getnode(int u,int v,int d){    int f1=top[u];    int f2=top[v];    while(f1!=f2)    {        if(deep[f1]<deep[f2])        {            swap(f1,f2);            swap(u,v);        }        g1[p[f1]]+=d;        g1[p[u]+1]-=d;        u=fa[f1];        f1=top[u];    }    if(u==v)    {        g1[p[u]]+=d;        g1[p[u]+1]-=d;        return;    }    if(deep[u]>deep[v])swap(u,v);    g1[p[u]]+=d;    g1[p[v]+1]-=d;    return;}void getedge(int u,int v,int d){    int f1=top[u];    int f2=top[v];    while(f1!=f2)    {        if(deep[f1]<deep[f2])        {            swap(f1,f2);            swap(u,v);        }        g2[p[f1]]+=d;        g2[p[u]+1]-=d;        u=fa[f1];        f1=top[u];    }    if(u==v)        return;    if(deep[u]>deep[v])swap(u,v);    g2[p[son[u]]]+=d;    g2[p[v]+1]-=d;    return;}struct lede{    int u,v;}e[M];int main(){    int T,m,n,i,u,v,w,kk=1;    char ch[22];    cin>>T;    while(T--)    {        scanf("%d%d",&n,&m);        for(i=0;i<=n;i++)            edge[i].clear();        for(i=1;i<n;i++)        {            scanf("%d%d",&u,&v);            edge[u].push_back(v);            edge[v].push_back(u);            e[i].u=u;            e[i].v=v;        }        init();        memset(g1,0,sizeof(g1));        memset(g2,0,sizeof(g2));        while(m--)        {            scanf("%s%d%d%d",ch,&u,&v,&w);            if(strcmp(ch,"ADD1")==0)                getnode(u,v,w);            else                getedge(u,v,w);        }        printf("Case #%d:\n",kk++);        int sum=0;        for(i=0;i<pos;i++)        {            sum+=g1[i];            ans[i]=sum;        }        for(i=1;i<=n;i++)        {            if(i==1)                printf("%d",ans[p[i]]);            else                printf(" %d",ans[p[i]]);        }        printf("\n");        sum=0;        for(i=1;i<pos;i++)        {            sum+=g2[i];            ans[i]=sum;        }        for(i=1;i<n;i++)        {            if(deep[e[i].u]<deep[e[i].v])                swap(e[i].u,e[i].v);            if(i==1)                printf("%d",ans[p[e[i].u]]);            else                printf(" %d",ans[p[e[i].u]]);        }        printf("\n");    }}