GSS5 spoj 2916. Can you answer these queries V 线段树
gss5 Can you answer these queries V
给出数列a1...an，询问时给出：
Query(x1,y1,x2,y2) = Max { A[i]+A[i+1]+...+A[j] ; x1 <= i <= y1 , x2 j <= y2 and x1 <= x2 , y1 <= y2 }
分析：
其实画个图分类讨论一下之后，跟gss1基本一样。。。
注意到x1<=x2 , y1<=y2.
所以大致可以分为：
1.y1<x2:
直接计算区间[x1,y1]的右子区间连续最大和，[x2,y2]的左区间连续最大和，如果y1与x2之间有空隙的话，需要加上[y1+1,x2-1]的和。
2.y2>=x2:
考虑x1与x2的关系：
如果x1==x2，最大值可能出现在区间[x1,y1]的最大子段和，[x1,y1]的右区间连续最大和加上[y1+1,y2]的左区间连续最大和。
否则，考虑三个区间[x1,x2-1],[x2,y1],[y1+1,y2]，这时考虑方式跟上面差不多，就不写出来了，具体可以看代码。
#include <set>
#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define debug puts("here")
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define foreach(i,vec) for(unsigned i=0;i<vec.size();i++)
#define pb push_back
#define RD(n) scanf("%d",&n)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define RD4(x,y,z,w) scanf("%d%d%d%d",&x,&y,&z,&w)
#define All(vec) vec.begin(),vec.end()
#define MP make_pair
#define PII pair<int,int>
#define PQ priority_queue
#define cmax(x,y) x = max(x,y)
#define cmin(x,y) x = min(x,y)
#define Clear(x) memset(x,0,sizeof(x))
/*
#pragma comment(linker, "/STACK:1024000000,1024000000")
int size = 256 << 20; // 256MB
char *p = (char*)malloc(size) + size;
__asm__("movl %0, %%esp\n" :: "r"(p) );
*/
/******** program ********************/
const int MAXN = 100005;
int a[MAXN];
struct segTree{
int l,r,lx,rx,mx,sum;
inline int mid(){
return (l+r)>>1;
}
}tree[MAXN<<2];
inline void Union(segTree& now,segTree l,segTree r){
now.lx = max( l.lx , l.sum+max(0,r.lx) );
now.rx = max( r.rx , r.sum+max(0,l.rx) );
now.mx = max( l.rx+r.lx , max(l.mx,r.mx) );
now.sum = l.sum+r.sum;
}
void build(int l,int r,int rt){
tree[rt].l = l;
tree[rt].r = r;
if(l==r){
tree[rt].lx = tree[rt].rx = tree[rt].sum = tree[rt].mx = a[l];
return;
}
int mid = tree[rt].mid();
build(l,mid,rt<<1);
build(mid+1,r,rt<<1|1);
Union(tree[rt],tree[rt<<1],tree[rt<<1|1]);
}
void modify(int pos,int c,int rt){
if(tree[rt].l==tree[rt].r){
tree[rt].lx = tree[rt].rx = tree[rt].mx = tree[rt].sum = c;
return;
}
int mid = tree[rt].mid();
if(pos<=mid)modify(pos,c,rt<<1);
else modify(pos,c,rt<<1|1);
Union(tree[rt],tree[rt<<1],tree[rt<<1|1]);
}
segTree ask(int l,int r,int rt){
if(l<=tree[rt].l&&r>=tree[rt].r)
return tree[rt];
int mid = tree[rt].mid();
segTree ans;
if(r<=mid) ans = ask(l,r,rt<<1);
else if(l>mid) ans = ask(l,r,rt<<1|1);
else{
segTree a = ask(l,r,rt<<1);
segTree b = ask(l,r,rt<<1|1);
Union( ans,a,b );
}
Union(tree[rt],tree[rt<<1],tree[rt<<1|1]);
return ans;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("sum.in","r",stdin);
//freopen("sum.out","w",stdout);
#endif
int m,n,x,y,c,d;
int ncase;
RD(ncase);
while(ncase--){
RD(n);
rep1(i,n)
RD(a[i]);
build(1,n,1);
RD(m);
segTree ca,cb,cc;
while(m--){
RD4(x,y,c,d);
int sum = 0;
if(y>=c){
if(x<c){
ca = ask(x,c-1,1);
cb = ask(c,y,1);
sum = max( ca.rx+cb.lx,cb.mx );
if(y<d){
cc = ask(y+1,d,1);
int tmp = max( max(0,ca.rx)+cb.sum+cc.lx , cb.rx+cc.lx );
sum = max( sum,tmp );
}
}else{
ca = ask(c,y,1);
sum = ca.mx;
if(y<d){
cc = ask(y+1,d,1);
sum = max(sum,ca.rx+cc.lx);
}
}
}else{
ca = ask(x,y,1);
cb = ask(c,d,1);
sum = ca.rx+cb.lx;
if(y+1<c){
cc = ask(y+1,c-1,1);
sum += cc.sum;
}
}
printf("%d\n",sum);
}
}
return 0;
}