HDU-3966 Aragorn's Story(树链剖分+线段树)

Aragorn's Story

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12479    Accepted Submission(s): 3331

Problem Description

Our protagonist is the handsome human prince Aragorn comes from The Lord of the Rings. One day Aragorn finds a lot of enemies who want to invade his kingdom. As Aragorn knows, the enemy has N camps out of his kingdom and M edges connect them. It is guaranteed that for any two camps, there is one and only one path connect them. At first Aragorn know the number of enemies in every camp. But the enemy is cunning , they will increase or decrease the number of soldiers in camps. Every time the enemy change the number of soldiers, they will set two camps C1 and C2. Then, for C1, C2 and all camps on the path from C1 to C2, they will increase or decrease K soldiers to these camps. Now Aragorn wants to know the number of soldiers in some particular camps real-time.

 

 

Input

Multiple test cases, process to the end of input.

For each case, The first line contains three integers N, M, P which means there will be N(1 ≤ N ≤ 50000) camps, M(M = N-1) edges and P(1 ≤ P ≤ 100000) operations. The number of camps starts from 1.

The next line contains N integers A1, A2, ...AN(0 ≤ Ai ≤ 1000), means at first in camp-i has Ai enemies.

The next M lines contains two integers u and v for each, denotes that there is an edge connects camp-u and camp-v.

The next P lines will start with a capital letter 'I', 'D' or 'Q' for each line.

'I', followed by three integers C1, C2 and K( 0≤K≤1000), which means for camp C1, C2 and all camps on the path from C1 to C2, increase K soldiers to these camps.

'D', followed by three integers C1, C2 and K( 0≤K≤1000), which means for camp C1, C2 and all camps on the path from C1 to C2, decrease K soldiers to these camps.

'Q', followed by one integer C, which is a query and means Aragorn wants to know the number of enemies in camp C at that time.

 

 

Output

For each query, you need to output the actually number of enemies in the specified camp.

 

 

Sample Input

3 2 5 1 2 3 2 1 2 3 I 1 3 5 Q 2 D 1 2 2 Q 1 Q 3

 

 

Sample Output

7 4 8

Hint

1.The number of enemies may be negative. 2.Huge input, be careful.

 

 

Source

2011 Multi-University Training Contest 13 - Host by HIT

 

 

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HDU是真鸡巴垃圾的一个OJ  老子的程序明明没有什么问题  结果判我RE

今天学了树链剖分  简单来说就是把树上的节点标个号  然后拆成一条条链来操作，这一条条链的一个重要的性质是他们的编号都是连在一起的  这些操作比如区间修改查询等需要依托线段树，splay等等数据结构来实现

好吧这是目前为止写的最长的程序。。。。。

#pragma comment(linker, "/STACK:1024000000,1024000000")  //据说这玩意可以手动扩栈
#include "bits/stdc++.h"
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long LL;
const int MAX=50005;
int n,m,T;
int tot,ttt;
int head[MAX],adj[MAX<<2],nex[MAX<<2];
int a[MAX],deep[MAX],fa[MAX];
int siz[MAX],son[MAX],top[MAX],tid[MAX],rak[MAX];
void addedge(int u,int v){
    tot++;
    adj[tot]=v;
    nex[tot]=head[u];
    head[u]=tot;
}
void init(){
    int i,j;
    mem(head,0),mem(son,0);
    tot=ttt=0;
}
void dfs1(int x,int ff,int de){ //标父亲,后代个数(包括自己),深度,重儿子 
    int i,j;
    deep[x]=de;
    siz[x]=1;fa[x]=ff;
    for (i=head[x];i;i=nex[i]){
        int v=adj[i];
        if (v==ff) continue;
        dfs1(v,x,de+1);
        siz[x]+=siz[v];
        if (!son[x] || siz[v]>siz[son[x]]){
            son[x]=v;
        }
    }
}
void dfs2(int x,int tp){//编号,拆树 
    int i,j;
    top[x]=tp;
    tid[x]=++ttt;
    rak[tid[ttt]]=x;
    if (!son[x]) return;
    dfs2(son[x],tp);
    for (i=head[x];i;i=nex[i]){
        if (adj[i]!=son[x] && adj[i]!=fa[x]){
            dfs2(adj[i],adj[i]);
        }
    }
}
//线段树 
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
int sum[MAX<<2],la[MAX<<2];

void PushUp(int rt){
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void PushDown(int rt,int l,int r){
    if (la[rt]){
        int m=(l+r)>>1;
        la[rt<<1]+=la[rt];la[rt<<1|1]+=la[rt];
        sum[rt<<1]+=(m-l+1)*la[rt];
        sum[rt<<1|1]+=(r-m)*la[rt];
        la[rt]=0;
    }
}
void build(int rt,int l,int r){
    la[rt]=0;
    if(l==r){
        sum[rt]=a[rak[l]];
        return;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    PushUp(rt);
}
void update(int rt,int l,int r,int x,int y,int z){
    if (x<=l && r<=y){
        sum[rt]+=z*(r-l+1);
        la[rt]+=z;
        return;
    }
    int m=(l+r)>>1;
    PushDown(rt,l,r);
    if (x<=m) update(lson,x,y,z);
    if (y>m) update(rson,x,y,z);
    PushUp(rt);
}
int search(int rt,int l,int r,int x){
    if (l==r){
        return sum[rt];
    }
    int res=0;
    int m=(l+r)>>1;
    PushDown(rt,l,r);
    if (x<=m) res+=search(lson,x);
    if (x>m) res+=search(rson,x);
    PushUp(rt);
    return res;
}
void calc(int x,int y,int z){
    int i,j;
    while (top[x]!=top[y]){
        if (deep[top[x]]<deep[top[y]]) swap(x,y);
        update(1,1,n,tid[top[x]],tid[x],z);
        x=fa[top[x]];
    }
    if (deep[x]>deep[y]) swap(x,y);
    update(1,1,n,tid[x],tid[y],z);
}
int main(){
    freopen ("story.in","r",stdin);
    freopen ("story.out","w",stdout);
    int i,j;
    int u,v,x,y,z;
    char c;
    while (~scanf("%d%d%d",&n,&m,&T)){
        init();
        for (i=1;i<=n;i++){
            scanf("%d",a+i);
        }
        for (i=1;i<=m;i++){
            scanf("%d%d\n",&u,&v);
            addedge(u,v);
            addedge(v,u);
        }
        dfs1(1,0,1);
        dfs2(1,1);
        build(1,1,n);
        while (T--){
            c=getchar();//cout<<c<<endl;
            if (c=='Q'){
                scanf("%d\n",&x);
                printf("%d\n",search(1,1,n,tid[x]));
            }
            else{
                scanf("%d%d%d\n",&x,&y,&z);
                if (c=='D') z=-z;
                calc(x,y,z);
            }
        }
    }
    return 0;
}