HDU-3966 Aragorn's Story(树链剖分+线段树)
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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.
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
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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。。。。。
1 #pragma comment(linker, "/STACK:1024000000,1024000000") //据说这玩意可以手动扩栈 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){ //标父亲,后代个数(包括自己),深度,重儿子 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){//编号,拆树 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 //线段树 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 }
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