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Aragorn's Story

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3494    Accepted Submission(s): 973


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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  1 #pragma comment(linker,"/STACK:100000000,100000000")
  2 #include<iostream>
  3 #include<stdio.h>
  4 #include<cstring>
  5 #include<cstdlib>
  6 using namespace std;
  7 typedef __int64 LL;
  8 const int maxn = 5e4+3;
  9 
 10 int pos,cont;
 11 int a[maxn];
 12 int son[maxn];
 13 int head[maxn];
 14 int vis[maxn];
 15 int w[maxn];
 16 int dep[maxn];
 17 int father[maxn];
 18 int hxl[maxn];
 19 int top[maxn];
 20 struct Edge
 21 {
 22     int to;
 23     int next;
 24 }edge[maxn*2];
 25 
 26 void init()
 27 {
 28     pos = cont = 0;
 29     memset(son,-1,sizeof(son));
 30     memset(head,-1,sizeof(head));
 31     memset(hxl,0,sizeof(hxl));
 32 }
 33 void addedge(int u,int v)
 34 {
 35     edge[cont].to = v;
 36     edge[cont].next = head[u];
 37     head[u] = cont;
 38     ++cont;
 39 }
 40 void dfs1(int u,int fre,int deep)
 41 {
 42     father[u] = fre;
 43     dep[u] = deep;
 44     vis[u] = 1;
 45     for(int i=head[u];i!=-1;i=edge[i].next)
 46     {
 47         int v = edge[i].to;
 48         if(v!=fre)
 49         {
 50             dfs1(v,u,deep+1);
 51             vis[u]=vis[u]+vis[v];
 52             if(son[u] == -1 || vis[v] > vis[son[u]])
 53             son[u] = v;
 54         }
 55     }
 56 }
 57 void dfs2(int u,int t)
 58 {
 59     top[u] = t;
 60     if(son[u]!=-1)
 61     {
 62         w[u]=++pos;
 63         dfs2(son[u],t);
 64     }
 65     else
 66     {
 67         w[u]=++pos;
 68         return;
 69     }
 70     for(int i=head[u];i!=-1;i=edge[i].next)
 71     {
 72         int v = edge[i].to;
 73         if(v!=son[u] && v!=father[u])
 74         dfs2(v,v);
 75     }
 76 }
 77 void add(int x,int n,int num1)
 78 {
 79     for(int i=x;i<=n;i=i+(i&(-i)))
 80     hxl[i] = hxl[i] + num1;
 81 }
 82 LL query(int x)
 83 {
 84     if(x==0)return 0;
 85     LL sum1 = 0;
 86     while(x)
 87     {
 88         sum1=sum1+hxl[x];
 89         x=x-(x&(-x));
 90     }
 91     return sum1;
 92 }
 93 void insert(int u,int v,int num1,int size1)
 94 {
 95     int topu=top[u],topv=top[v];
 96     while(topu!=topv)
 97     {
 98         if(dep[topu]<dep[topv])
 99         {
100             swap(u,v);
101             swap(topu,topv);
102         }
103         add(w[topu],pos,num1*size1);
104         add(w[u]+1,pos,-1*num1*size1);
105         u=father[topu];
106         topu = top[u];
107     }
108     if(dep[u]>dep[v]) swap(u,v);
109     add(w[u],pos,num1*size1);
110     add(w[v]+1,pos,-1*num1*size1);
111 }
112 int main()
113 {
114     int n,m,q,u,v,l,r;
115     while(scanf("%d%d%d",&n,&m,&q)>0)
116     {
117         init();
118         for(int i=1;i<=n;i++)
119         scanf("%d",&a[i]);
120         for(int i=1;i<=m;i++)
121         {
122             scanf("%d%d",&u,&v);
123             addedge(u,v);
124             addedge(v,u);
125         }
126         addedge(0,1);
127         addedge(1,0);
128         dfs1(1,0,0);
129         dfs2(1,1);
130         char str1[6];
131         while(q--)
132         {
133             scanf("%s",str1);
134             if(str1[0]=='I')
135             {
136                 scanf("%d%d%d",&l,&r,&v);
137                 insert(l,r,v,1);
138             }
139             else if(str1[0]=='D')
140             {
141                 scanf("%d%d%d",&l,&r,&v);
142                 insert(l,r,v,-1);
143             }
144             else if(str1[0]=='Q')
145             {
146                 scanf("%d",&r);
147                 printf("%I64d\n",query(w[r])+a[r]);
148             }
149         }
150     }
151     return 0;
152 }

 

posted @ 2014-11-04 09:25  芷水  阅读(145)  评论(0)  编辑  收藏  举报