real letter 們的 等等 then 需要 family sea inpu

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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  1 #pragma comment(linker, "/STACK:1024000000,1024000000")  //據說這玩意可以手動擴棧
  2 #include <cstdio>
  3 #include <cmath>
  4 #include <cstring>
  5 #include <cstdlib>
  6 #include <queue>
  7 #include <stack>
  8 #include <vector>
  9 #include <iostream>
 10 #include "algorithm"
 11 #define mem(a,b) memset(a,b,sizeof(a))
 12 using namespace std;
 13 typedef long long LL;
 14 const int MAX=50005;
 15 int n,m,T;
 16 int tot,ttt;
 17 int head[MAX],adj[MAX<<2],nex[MAX<<2];
 18 int a[MAX],deep[MAX],fa[MAX];
 19 int siz[MAX],son[MAX],top[MAX],tid[MAX],rak[MAX];
 20 void addedge(int u,int v){
 21     tot++;
 22     adj[tot]=v;
 23     nex[tot]=head[u];
 24     head[u]=tot;
 25 }
 26 void init(){
 27     int i,j;
 28     mem(head,0),mem(son,0);
 29     tot=ttt=0;
 30 }
 31 void dfs1(int x,int ff,int de){ //標父親,後代個數(包括自己),深度,重兒子 
 32     int i,j;
 33     deep[x]=de;
 34     siz[x]=1;fa[x]=ff;
 35     for (i=head[x];i;i=nex[i]){
 36         int v=adj[i];
 37         if (v==ff) continue;
 38         dfs1(v,x,de+1);
 39         siz[x]+=siz[v];
 40         if (!son[x] || siz[v]>siz[son[x]]){
 41             son[x]=v;
 42         }
 43     }
 44 }
 45 void dfs2(int x,int tp){//編號,拆樹 
 46     int i,j;
 47     top[x]=tp;
 48     tid[x]=++ttt;
 49     rak[tid[ttt]]=x;
 50     if (!son[x]) return;
 51     dfs2(son[x],tp);
 52     for (i=head[x];i;i=nex[i]){
 53         if (adj[i]!=son[x] && adj[i]!=fa[x]){
 54             dfs2(adj[i],adj[i]);
 55         }
 56     }
 57 }
 58 //線段樹 
 59 #define lson rt<<1,l,m
 60 #define rson rt<<1|1,m+1,r
 61 int sum[MAX<<2],la[MAX<<2];
 62 
 63 void PushUp(int rt){
 64     sum[rt]=sum[rt<<1]+sum[rt<<1|1];
 65 }
 66 void PushDown(int rt,int l,int r){
 67     if (la[rt]){
 68         int m=(l+r)>>1;
 69         la[rt<<1]+=la[rt];la[rt<<1|1]+=la[rt];
 70         sum[rt<<1]+=(m-l+1)*la[rt];
 71         sum[rt<<1|1]+=(r-m)*la[rt];
 72         la[rt]=0;
 73     }
 74 }
 75 void build(int rt,int l,int r){
 76     la[rt]=0;
 77     if(l==r){
 78         sum[rt]=a[rak[l]];
 79         return;
 80     }
 81     int m=(l+r)>>1;
 82     build(lson);
 83     build(rson);
 84     PushUp(rt);
 85 }
 86 void update(int rt,int l,int r,int x,int y,int z){
 87     if (x<=l && r<=y){
 88         sum[rt]+=z*(r-l+1);
 89         la[rt]+=z;
 90         return;
 91     }
 92     int m=(l+r)>>1;
 93     PushDown(rt,l,r);
 94     if (x<=m) update(lson,x,y,z);
 95     if (y>m) update(rson,x,y,z);
 96     PushUp(rt);
 97 }
 98 int search(int rt,int l,int r,int x){
 99     if (l==r){
100         return sum[rt];
101     }
102     int res=0;
103     int m=(l+r)>>1;
104     PushDown(rt,l,r);
105     if (x<=m) res+=search(lson,x);
106     if (x>m) res+=search(rson,x);
107     PushUp(rt);
108     return res;
109 }
110 void calc(int x,int y,int z){
111     int i,j;
112     while (top[x]!=top[y]){
113         if (deep[x]<deep[y]) swap(x,y);
114         update(1,1,n,tid[top[x]],tid[x],z);
115         x=fa[top[x]];
116     }
117     if (deep[x]>deep[y]) swap(x,y);
118     update(1,1,n,tid[x],tid[y],z);
119 }
120 int main(){
121     freopen ("story.in","r",stdin);
122     freopen ("story.out","w",stdout);
123     int i,j;
124     int u,v,x,y,z;
125     char c;
126     while (~scanf("%d%d%d",&n,&m,&T)){
127         init();
128         for (i=1;i<=n;i++){
129             scanf("%d",a+i);
130         }
131         for (i=1;i<=m;i++){
132             scanf("%d%d\n",&u,&v);
133             addedge(u,v);
134             addedge(v,u);
135         }
136         dfs1(1,0,1);
137         dfs2(1,1);
138         build(1,1,n);
139         while (T--){
140             c=getchar();//cout<<c<<endl;
141             if (c==‘Q‘){
142                 scanf("%d\n",&x);
143                 printf("%d\n",search(1,1,n,tid[x]));
144             }
145             else{
146                 scanf("%d%d%d\n",&x,&y,&z);
147                 if (c==‘D‘) z=-z;
148                 calc(x,y,z);
149             }
150         }
151     }
152     return 0;
153 }

HDU-3966 Aragorn's Story(樹鏈剖分+線段樹)