Aragorn's Story
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13587 Accepted Submission(s): 3623
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.
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
基于点权 单点查询 修改路径上的点权 模板
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<map> 6 using namespace std; 7 #define ll long long 8 #define mod 998244353 9 const int N=50010; 10 const int INF=0x3f3f3f3f; 11 struct Edge 12 { 13 int to,next; 14 } edge[2*N]; 15 int head[N]; 16 int top[N]; 17 int fa[N]; 18 int deep[N]; 19 int num[N]; 20 int p[N]; 21 int fp[N]; 22 int son[N]; 23 int pos; 24 int tot; 25 void init() 26 { 27 tot=0; 28 memset(head,-1,sizeof(head)); 29 pos=1; 30 memset(son,-1,sizeof(son)); 31 } 32 void addedge(int u,int v) 33 { 34 edge[tot].to=v; 35 edge[tot].next=head[u]; 36 head[u]=tot++; 37 } 38 void dfs1(int u,int pre,int d) 39 { 40 deep[u]=d; 41 fa[u]=pre; 42 num[u]=1; 43 for(int i=head[u]; i!=-1; i=edge[i].next) 44 { 45 int v=edge[i].to; 46 if(v!=pre) 47 { 48 dfs1(v,u,d+1); 49 num[u]+=num[v]; 50 if(son[u]==-1||num[v]>num[son[u]]) 51 son[u]=v; 52 } 53 } 54 } 55 void getpos(int u,int sp) 56 { 57 top[u]=sp; 58 p[u]=pos++; 59 fp[p[u]]=u; 60 if(son[u]==-1) return ; 61 getpos(son[u],sp); 62 for(int i=head[u]; i!=-1; i=edge[i].next) 63 { 64 int v=edge[i].to; 65 if(v!=son[u]&&v!=fa[u]) 66 getpos(v,v); 67 } 68 } 69 70 int lowbit(int x) 71 { 72 return x&(-x); 73 } 74 int c[N]; 75 int n; 76 int sum(int i) 77 { 78 int s=0; 79 while(i>0) 80 { 81 s+=c[i]; 82 i-=lowbit(i); 83 } 84 return s; 85 } 86 void add(int i,int val) 87 { 88 while(i<=n) 89 { 90 c[i]+=val; 91 i+=lowbit(i); 92 } 93 } 94 void change(int u,int v,int val) 95 { 96 int f1=top[u],f2=top[v]; 97 int tmp=0; 98 while(f1!=f2) 99 { 100 if(deep[f1]<deep[f2]) 101 { 102 swap(f1,f2); 103 swap(u,v); 104 } 105 add(p[f1],val); 106 add(p[u]+1,-val); 107 u=fa[f1]; 108 f1=top[u]; 109 } 110 if(deep[u]>deep[v]) swap(u,v); 111 add(p[u],val); 112 add(p[v]+1,-val); 113 } 114 int a[N]; 115 int main() 116 { 117 int M,P; 118 while(scanf("%d %d %d",&n,&M,&P)!=EOF) 119 { 120 int u,v; 121 int C1,C2,K; 122 char op[10]; 123 init(); 124 for(int i=1; i<=n; i++) 125 scanf("%d",&a[i]); 126 while(M--) 127 { 128 scanf("%d %d",&u,&v); 129 addedge(u,v); 130 addedge(v,u); 131 } 132 133 dfs1(1,0,0); 134 getpos(1,1); 135 memset(c,0,sizeof(c)); 136 for(int i=1; i<=n; i++) 137 { 138 add(p[i],a[i]); 139 add(p[i]+1,-a[i]); 140 } 141 142 while(P--) 143 { 144 scanf("%s",op); 145 if(op[0]=='Q') 146 { 147 scanf("%d",&u); 148 printf("%d ",sum(p[u])); 149 } 150 else 151 { 152 scanf("%d%d%d",&C1,&C2,&K); 153 if(op[0]=='D') 154 K=-K; 155 change(C1,C2,K); 156 } 157 } 158 } 159 return 0; 160 }