SPOJ GSS5 Can you answer these queries V（区间合并）

Can you answer these queries V

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:26 3 -2 1 -4 5 221 1 2 31 3 2 51 111 1 1 1Output:231

Hint

Added by:Frank Rafael ArteagaDate:2008-08-06Time limit:0.132sSource limit:50000BMemory limit:1536MBCluster:Cube (Intel G860)Languages:All except: C99 strict ERL JS NODEJS PERL 6 VB.netResource:K.-Y. Chen and K.-M. Chao, On the Range Maximum-Sum Segment Query Problem, 2007.

解题思路：

分为三种情况.

1. x1 y1 x2 y2  

这种情况就是   x1 y1 的右最大  +   sum【y1  x2】 + x2 y2的做最大

2.x1 x2 y2 y1 其实就是 y1==y2的时候

要么区间在 x2-y2之间

要么区间的头在  x1 x2之间，尾在 x2 y2之间

3. x1 x2 y1 y2。

这种情况和2比较像是  这个时候  x1-y2 区间分成三段。只需要枚举头尾各在区间哪个区间就好了。

需要注意的是：有可能 x2==y1。

所以要注意一下比较的时候的边界。

AC代码：

#include <iostream>#include <cstdio>#define lson id<<1,s,mid#define rson id<<1|1,mid+1,eusing namespace std;const int N = 10005;int tree[N<<2];int lef[N<<2];int rig[N<<2];int sum[N<<2];void pushup(int id){    sum[id] = sum[id<<1]+sum[id<<1|1];    lef[id] = max(lef[id<<1],sum[id<<1]+lef[id<<1|1]);    rig[id] = max(rig[id<<1|1],sum[id<<1|1]+rig[id<<1]);    tree[id] = max(tree[id<<1],max(tree[id<<1|1],lef[id<<1|1]+rig[id<<1]));}void build(int id,int s,int e){    if(s == e){        scanf("%d",&tree[id]);        lef[id] = rig[id] = sum[id] = tree[id];        return;    }    int mid = (s+e)>>1;    build(lson);    build(rson);    pushup(id);}int Q_sum(int id,int s,int e,int l,int r){    if(l > r)        return 0;    if(l<=s && r>=e){        return sum[id];    }    int mid = (s+e)>>1;    if(r <= mid)        return Q_sum(lson,l,r);    else if(l>mid)        return Q_sum(rson,l,r);    else{        return Q_sum(lson,l,mid)+Q_sum(rson,mid+1,r);    }}int Q_L(int id,int s,int e,int l,int r){    if(l > r)        return 0;    if(l<=s && r>=e){        return lef[id];    }    int mid = (s+e)>>1;    if(r <= mid)        return Q_L(lson,l,r);    else if(l > mid)        return Q_L(rson,l,r);    else return max(Q_L(lson,l,mid),max(Q_sum(lson,l,mid),Q_sum(lson,l,mid)+Q_L(rson,mid+1,r)));}int Q_R(int id,int s,int e,int l,int r){    if(l > r)        return 0;    if(l<=s && r>=e){        return rig[id];    }    int mid = (s+e)>>1;    if(r <= mid)        return Q_R(lson,l,r);    else if(l > mid)        return Q_R(rson,l,r);    else return max(Q_R(rson,mid+1,r),max(Q_sum(rson,mid+1,r),Q_sum(rson,mid+1,r)+Q_R(lson,l,mid)));}int query(int id,int s,int e,int l,int r){    if(l > r)        return 0;    if(l<=s && r>=e){        return tree[id];    }    int mid = (s+e)>>1;    if(r <= mid)        return query(lson,l,r);    else if(l > mid)        return query(rson,l,r);    else{        return max(Q_L(rson,mid+1,r)+Q_R(lson,l,mid),max(query(lson,l,mid),query(rson,mid+1,r)));    }}int main(){    int T;    scanf("%d",&T);    while(T--){        int n;        scanf("%d",&n);        build(1,1,n);        int m;        scanf("%d",&m);        while(m--){            int x1,x2,y1,y2;            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);            if(x2 > y1){                printf("%d\n",Q_R(1,1,n,x1,y1)+Q_sum(1,1,n,y1+1,x2-1)+Q_L(1,1,n,x2,y2));            }else if(y1 == y2){                printf("%d\n",max(query(1,1,n,x2,y2),Q_L(1,1,n,x2,y2)+Q_R(1,1,n,x1,x2-1)));            }else{                printf("%d\n",max(Q_R(1,1,n,x1,x2-1)+Q_L(1,1,n,x2,y1),max(Q_R(1,1,n,x1,x2-1)+Q_L(1,1,n,y1+1,y2)+Q_sum(1,1,n,x2,y1),max(Q_R(1,1,n,x2,y1)+Q_L(1,1,n,y1+1,y2),query(1,1,n,x2,y1)))));            }        }    }    return 0;}