地址：http://acm.split.hdu.edu.cn/showproblem.php?pid=6162
题目：
Ch’s gift
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 526 Accepted Submission(s): 177
Problem Description
Mr. Cui is working off-campus and he misses his girl friend very much. After a whole night tossing and turning, he decides to get to his girl friend's city and of course, with well-chosen gifts. He knows neither too low the price could a gift be since his girl friend won't like it, nor too high of it since he might consider not worth to do. So he will only buy gifts whose price is between [a,b].
There are n cities in the country and (n-1) bi-directional roads. Each city can be reached from any other city. In the ith city, there is a specialty of price ci Cui could buy as a gift. Cui buy at most 1 gift in a city. Cui starts his trip from city s and his girl friend is in city t. As mentioned above, Cui is so hurry that he will choose the quickest way to his girl friend(in other words, he won't pass a city twice) and of course, buy as many as gifts as possible. Now he wants to know, how much money does he need to prepare for all the gifts?
There are n cities in the country and (n-1) bi-directional roads. Each city can be reached from any other city. In the ith city, there is a specialty of price ci Cui could buy as a gift. Cui buy at most 1 gift in a city. Cui starts his trip from city s and his girl friend is in city t. As mentioned above, Cui is so hurry that he will choose the quickest way to his girl friend(in other words, he won't pass a city twice) and of course, buy as many as gifts as possible. Now he wants to know, how much money does he need to prepare for all the gifts?
Input
There are multiple cases.
For each case:
The first line contains tow integers n,m(1≤n,m≤10^5), representing the number of cities and the number of situations.
The second line contains n integers c1,c2,...,cn(1≤ci≤10^9), indicating the price of city i's specialty.
Then n-1 lines follows. Each line has two integers x,y(1≤x,y≤n), meaning there is road between city x and city y.
Next m line follows. In each line there are four integers s,t,a,b(1≤s,t≤n;1≤a≤b≤10^9), which indicates start city, end city, lower bound of the price, upper bound of the price, respectively, as the exact meaning mentioned in the description above
For each case:
The first line contains tow integers n,m(1≤n,m≤10^5), representing the number of cities and the number of situations.
The second line contains n integers c1,c2,...,cn(1≤ci≤10^9), indicating the price of city i's specialty.
Then n-1 lines follows. Each line has two integers x,y(1≤x,y≤n), meaning there is road between city x and city y.
Next m line follows. In each line there are four integers s,t,a,b(1≤s,t≤n;1≤a≤b≤10^9), which indicates start city, end city, lower bound of the price, upper bound of the price, respectively, as the exact meaning mentioned in the description above
Output
Output m space-separated integers in one line, and the ith number should be the answer to the ith situation.
Sample Input
5 3
1 2 1 3 2
1 2
2 4
3 1
2 5
4 5 1 3
1 1 1 1
3 5 2 3
Sample Output
7 1 4
Source
思路：
裸的树链剖分套主席树。
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 #define MP make_pair 6 #define PB push_back 7 typedef long long LL; 8 typedef pair<int,int> PII; 9 const double eps=1e-8; 10 const double pi=acos(-1.0); 11 const int K=1e5+100; 12 const int mod=1e9+7; 13 14 vector<int>mp[K]; 15 int top[K],sz[K],fa[K],son[K],id[K],hid[K],deep[K]; 16 int cnt,n,q; 17 18 void dfs1(int x,int f) 19 { 20 sz[x]=1,fa[x]=f,son[x]=-1,deep[x]=deep[f]+1; 21 for(int i=0;i<mp[x].size();i++) 22 if(mp[x][i]!=f) 23 { 24 dfs1(mp[x][i],x); 25 sz[x]+=sz[mp[x][i]]; 26 if(son[x]==-1||sz[son[x]]<sz[mp[x][i]]) 27 son[x]=mp[x][i]; 28 } 29 } 30 void dfs2(int x,int f) ///每条边用深度大的节点的序号表示 31 { 32 top[x]=f,id[x]=++cnt,hid[id[x]]=x; 33 if(son[x]!=-1) dfs2(son[x],f); 34 for(int i=0;i<mp[x].size();i++) 35 if(mp[x][i]!=fa[x]&&mp[x][i]!=son[x]) 36 dfs2(mp[x][i],mp[x][i]); 37 } 38 39 int tot,ls[K*20],rs[K*20],rt[K*20]; 40 int a[K],b[K]; 41 LL sum[K*20]; 42 //sum[o]记录的是该节点区间内出现的数的和 43 //区间指的是将数离散化后的区间 44 void build(int &o,int l,int r) 45 { 46 o=++tot,sum[o]=0; 47 int mid=l+r>>1; 48 if(l!=r) 49 build(ls[o],l,mid),build(rs[o],mid+1,r); 50 } 51 void update(int &o,int l,int r,int last,int x) 52 { 53 o=++tot,sum[o]=sum[last]+b[x]; 54 ls[o]=ls[last],rs[o]=rs[last]; 55 if(l==r) return ; 56 int mid=l+r>>1; 57 if(x<=mid) update(ls[o],l,mid,ls[last],x); 58 else update(rs[o],mid+1,r,rs[last],x); 59 } 60 LL query(int lo,int ro,int l,int r,int k) 61 { 62 if(k<1) return 0; 63 if(r<=k) return sum[ro]-sum[lo]; 64 int mid=l+r>>1; 65 if(k<=mid) return query(ls[lo],ls[ro],l,mid,k); 66 return query(rs[lo],rs[ro],mid+1,r,k)+sum[ls[ro]]-sum[ls[lo]]; 67 } 68 LL tree_query(int x,int y,int l,int r,int sz) 69 { 70 LL ret=0; 71 while(top[x]!=top[y]) 72 { 73 if(deep[top[x]]<deep[top[y]]) swap(x,y); 74 ret+=query(rt[id[top[x]]-1],rt[id[x]],1,sz,r)-query(rt[id[top[x]]-1],rt[id[x]],1,sz,l-1); 75 x=fa[top[x]]; 76 } 77 if(deep[x]>deep[y]) swap(x,y); 78 ret+=query(rt[id[x]-1],rt[id[y]],1,sz,r)-query(rt[id[x]-1],rt[id[y]],1,sz,l-1); 79 return ret; 80 } 81 int main(void) 82 { 83 //freopen("in.acm","r",stdin); 84 //freopen("out.acm","w",stdout); 85 while(~scanf("%d%d",&n,&q)) 86 { 87 cnt=tot=0; 88 memset(mp,0,sizeof mp); 89 for(int i=1;i<=n;i++) scanf("%d",a+i),b[i]=a[i]; 90 for(int i=1,x,y;i<n;i++) 91 scanf("%d%d",&x,&y),mp[x].PB(y),mp[y].PB(x); 92 sort(b+1,b+1+n); 93 int sz=unique(b+1,b+1+n)-b-1; 94 for(int i=1;i<=n;i++) 95 a[i]=lower_bound(b+1,b+1+sz,a[i])-b; 96 dfs1(1,0); 97 dfs2(1,0); 98 build(rt[0],1,sz); 99 for(int i=1;i<=n;i++) 100 update(rt[i],1,sz,rt[i-1],a[hid[i]]); 101 // for(int i=1;i<=n;i++) 102 // printf("id[%d]=%d ",i,id[i]); 103 // printf(" "); 104 for(int i=1,u,v,l,r,tmp;i<=q;i++) 105 { 106 scanf("%d%d%d%d",&u,&v,&l,&r); 107 l=lower_bound(b+1,b+1+sz,l)-b; 108 tmp=lower_bound(b+1,b+1+sz,r)-b; 109 if(b[tmp]>r||tmp>sz) r=tmp-1; 110 else r=tmp; 111 printf("%lld%c",tree_query(u,v,l,r,sz),i==q?' ':' '); 112 } 113 } 114 return 0; 115 }