| Time Limit: 132MS | Memory Limit: 1572864KB | 64bit IO Format: %lld & %llu |
Description
You are given a sequence A[1], A[2], ..., A[N] . ( |A[i]| <= 10000 , 1 <= N <= 10000 ). A query is defined as follows: Query(x1,y1,x2,y2) = Max { A[i]+A[i+1]+...+A[j] ; x1 <= i <= y1 , x2 <= j <= y2 and x1 <= x2 , y1 <= y2 }. Given M queries (1 <= M <= 10000), your program must output the results of these queries.
Input
The first line of the input consist of the number of tests cases <= 5. Each case consist of the integer N and the sequence A. Then the integer M. M lines follow, contains 4 numbers x1, y1, x2 y2.
Output
Your program should output the results of the M queries for each test case, one query per line.
Example
Input: 2 6 3 -2 1 -4 5 2 2 1 1 2 3 1 3 2 5 1 1 1 1 1 1 1 Output: 2 3 1
Hint
| Added by: | Frank Rafael Arteaga |
| Date: | 2008-08-06 |
| Time limit: | 0.132s |
| Source limit: | 50000B |
| Memory limit: | 1536MB |
| Cluster: | Cube (Intel G860) |
| Languages: | All except: C99 strict ERL JS NODEJS PERL 6 VB.net |
| Resource: | K.-Y. Chen and K.-M. Chao, On the Range Maximum-Sum Segment Query Problem, 2007. |
又是查询最大连续字段和，但是限制了左右端点所在的区间……
线段树的部分不需要改动，计算答案的时候改一下即可。
如果区间有重复部分，就把区间分成三段，左段里找左端点，右段里找右端点，然后并上中段。有一串麻烦的判断，具体看代码。
如果区间没有重复部分，就左段里找左端点，右段里找右端点，然后强制加上两区间中间的序列和。
1 /*by SilverN*/ 2 #include<algorithm> 3 #include<iostream> 4 #include<cstring> 5 #include<cstdio> 6 #include<cmath> 7 #define lc rt<<1 8 #define rc rt<<1|1 9 using namespace std; 10 const int mxn=100010; 11 int read(){ 12 int x=0,f=1;char ch=getchar(); 13 while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();} 14 while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();} 15 return x*f; 16 } 17 int n,m; 18 int data[mxn]; 19 struct node{ 20 int mx; 21 int ml,mr; 22 int smm; 23 }t[mxn<<2],tmp0; 24 void Build(int l,int r,int rt){ 25 if(l==r){t[rt].mx=t[rt].ml=t[rt].mr=data[l];t[rt].smm=data[l];return;} 26 int mid=(l+r)>>1; 27 Build(l,mid,lc); 28 Build(mid+1,r,rc); 29 t[rt].smm=t[lc].smm+t[rc].smm; 30 t[rt].mx=max(t[lc].mx,t[rc].mx); 31 t[rt].mx=max(t[lc].mr+t[rc].ml,t[rt].mx); 32 t[rt].ml=max(t[lc].ml,t[lc].smm+t[rc].ml); 33 t[rt].mr=max(t[rc].mr,t[rc].smm+t[lc].mr); 34 return; 35 } 36 node query(int L,int R,int l,int r,int rt){ 37 // printf("%d %d %d %d %d ",L,R,l,r,rt); 38 if(R<L){ 39 return (node){0,0,0,0}; 40 } 41 if(L<=l && r<=R){return t[rt];} 42 int mid=(l+r)>>1; 43 node res1; 44 if(L<=mid)res1=query(L,R,l,mid,lc); 45 else res1=tmp0; 46 node res2; 47 if(R>mid)res2=query(L,R,mid+1,r,rc); 48 else res2=tmp0; 49 node res={0}; 50 res.smm=res1.smm+res2.smm; 51 res.mx=max(res1.mx,res2.mx); 52 res.mx=max(res.mx,res1.mr+res2.ml); 53 res.ml=max(res1.ml,res1.smm+res2.ml); 54 res.mr=max(res2.mr,res2.smm+res1.mr); 55 return res; 56 } 57 int qsum(int L,int R,int l,int r,int rt){ 58 if(L<=l && r<=R)return t[rt].smm; 59 int mid=(l+r)>>1; 60 int res=0; 61 if(L<=mid)res+=qsum(L,R,l,mid,lc); 62 if(R>mid)res+=qsum(L,R,mid+1,r,rc); 63 return res; 64 } 65 int main(){ 66 int T; 67 T=read(); 68 while(T--){ 69 n=read(); 70 int i,j,x0,y0,x2,y2; 71 for(i=1;i<=n;i++)data[i]=read(); 72 Build(1,n,1); 73 m=read(); 74 tmp0.ml=tmp0.mr=tmp0.mx=-1e9;tmp0.smm=0; 75 for(i=1;i<=m;i++){ 76 x0=read();y0=read();x2=read();y2=read(); 77 int tmp=0; 78 int ans=-1e9; 79 if(y0>=x2){ 80 //区间重叠 81 node res=query(x2,y0,1,n,1); 82 node res1=query(x0,x2-1,1,n,1); 83 node res2=query(y0+1,y2,1,n,1); 84 ans=max(ans,res1.mr+res.smm+res2.ml); 85 ans=max(ans,res1.mr+res.ml); 86 ans=max(ans,res.mr+res2.ml); 87 ans=max(ans,res.mx); 88 } 89 else{ 90 //区间未重叠 91 if(y0+1<x2)tmp=qsum(y0+1,x2-1,1,n,1); 92 node res1=query(x0,y0,1,n,1); 93 node res2=query(x2,y2,1,n,1); 94 ans=max(ans,tmp+res1.mr+res2.ml); 95 } 96 //printf("%d ",query(x2,y2,1,n,1).mx); 97 printf("%d ",ans); 98 } 99 } 100 return 0; 101 }