Description
After doing Ray a great favor to collect sticks for Ray, Poor Neal becomes very hungry. In return for Neal‘s help, Ray makes a great dinner for Neal. When it is time for dinner, Ray arranges all the dishes he
makes in a single line (actually this line is very long ... , the dishes are represented by 1, 2, 3 ... ). ``You make me work hard and don‘t pay me! You refuse to teach me Latin Dance! Now it is time for
you to serve me", Neal says to himself.
Every dish has its own value represented by an integer whose absolute value is less than 1,000,000,000. Before having dinner, Neal is wondering about the total value of the dishes he will eat. So he raises many
questions about the values of dishes he would have.
For each question Neal asks, he will first write down an interval [a, b] (inclusive) to represent all the dishes a, a + 1,..., b ,
where a and b are positive integers, and then asks Ray which sequence of consecutive dishes in the interval has the most total value. Now Ray needs your help.
Input
The input file contains multiple test cases. For each test case, there are two integers n and m in the first line (n, m <
500000) . n is the number of dishes and m is the number of questions Neal asks.
Then n numbers come in the second line, which are the values of the dishes from left to right. Next m lines are the questions and each line
contains two numbers a , b as described above. Proceed to the end of the input file.
Output
For each test case, output m lines. Each line contains two numbers, indicating the beginning position and end position of the sequence. If there are multiple solutions, output
the one with the smallest beginning position. If there are still multiple solutions then, just output the one with the smallest end position. Please output the result as in the Sample Output.
Sample Input
3 1 1 2 3 1 1
Sample Output
Case 1: 1 1
题意：
给出长度为n的整数序列，然后m次询问，对于每次询问，要求找到区间内的两个下标x,y使得区间和尽量大，如果有多解，x尽量小，还有多解，那么y也尽量小
思路：
线段树，建树过程中要标记
max_all:最大连续和
max_prefix:最大前缀和
max_suffix:最大后缀和
pre_r:最大前缀的结束位置
suf_l:最大后缀的开始位置
sum:区间总和
对于更新，由于要x尽量小与y尽量小，所以在更新的时候我们要确定好更新的顺序，也就是l先尽量小，然后才是y尽量小
更新有5种状况，那么按顺序就是
1.左边的sum+右区间的max_prefix;
2.左区间max_all
3.左边的max_suffix+右边的max_prefix
4.左边的max_suffix+右边的sum
5.右边的max_all
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;
#define lson 2*i
#define rson 2*i+1
#define LS l,mid,lson
#define RS mid+1,r,rson
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 500005
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define lowbit(x) (x&-x)
LL s[N];
int n,m;
struct node
{
int l,r,pre_r,suf_l;
LL max_all,max_prefix,max_suffix,sum;
} a[N<<2];
void pushup(int l,int r,int i)
{
int mid = (l+r)/2;
a[i].sum = a[lson].sum+a[rson].sum;
a[i].max_prefix = a[lson].max_prefix;
a[i].pre_r = a[lson].pre_r;
a[i].max_suffix = a[rson].max_suffix;
a[i].suf_l = a[rson].suf_l;
//1.左边的sum+右区间的max_prefix;
a[i].max_all = a[lson].sum+a[rson].max_prefix;
a[i].l = l;
a[i].r = a[rson].pre_r;
//2.左区间max_all
if(a[i].max_all<a[lson].max_all || (a[i].max_all==a[lson].max_all&&a[i].l==a[lson].l))
{
a[i].max_all = a[lson].max_all;
a[i].l = a[lson].l;
a[i].r = a[lson].r;
}
//3.左边的max_suffix+右边的max_prefix
if(a[i].max_all<a[lson].max_suffix+a[rson].max_prefix || (a[i].max_all==a[lson].max_suffix+a[rson].max_prefix &&a[i].l>a[lson].suf_l))
{
a[i].max_all=a[lson].max_suffix+a[rson].max_prefix;
a[i].l = a[lson].suf_l;
a[i].r = a[rson].pre_r;
}
//4.左边的max_suffix+右边的sum
if(a[i].max_all<a[lson].max_suffix+a[rson].sum || (a[i].max_all==a[lson].max_suffix+a[rson].sum &&a[i].l>a[lson].suf_l))
{
a[i].max_all=a[lson].max_suffix+a[rson].sum;
a[i].l = a[lson].suf_l;
a[i].r = r;
}
//5.右边的max_all
if(a[i].max_all<a[rson].max_all || (a[i].max_all==a[rson].max_all&&a[i].l>a[rson].suf_l))
{
a[i].max_all = a[rson].max_all;
a[i].l = a[rson].l;
a[i].r = a[rson].r;
}
//更新前缀与后缀的位置
if(a[lson].sum+a[rson].max_prefix>a[i].max_prefix)
{
a[i].max_prefix=a[lson].sum+a[rson].max_prefix;
a[i].pre_r = a[rson].pre_r;
}
if(a[rson].sum+a[lson].max_suffix>=a[i].max_suffix)
{
a[i].max_suffix=a[rson].sum+a[lson].max_suffix;
a[i].suf_l = a[lson].suf_l;
}
}
void init(int l,int r,int i)
{
if(l == r)
{
a[i].max_all = a[i].max_prefix = a[i].max_suffix = a[i].sum = s[l];
a[i].l = a[i].r = a[i].pre_r = a[i].suf_l = l;
return ;
}
int mid = (l+r)/2;
init(LS);
init(RS);
pushup(l,r,i);
}
node query(int i,int l,int r,int L,int R)
{
if(l==L&&r==R)
{
return a[i];
}
int mid=(L+R)>>1;
if(r<=mid) return query(lson,l,r,L,mid);
else if(l>mid) return query(rson,l,r,mid+1,R);
else
{
node t1=query(lson,l,mid,L,mid);
node t2=query(rson,mid+1,r,mid+1,R);
node t;
t.sum = t1.sum+t2.sum;
t.max_prefix = t1.max_prefix;
t.pre_r = t1.pre_r;
t.max_suffix = t2.max_suffix;
t.suf_l = t2.suf_l;
t.max_all = t1.sum+t2.max_prefix;
t.l = l;
t.r = t2.pre_r;
if(t.max_all<t1.max_all || (t.max_all==t1.max_all&&t.l==t1.l))
{
t.max_all = t1.max_all;
t.l = t1.l;
t.r = t1.r;
}
if(t.max_all<t1.max_suffix+t2.max_prefix || (t.max_all==t1.max_suffix+t2.max_prefix &&t.l>t1.suf_l))
{
t.max_all=t1.max_suffix+t2.max_prefix;
t.l = t1.suf_l;
t.r = t2.pre_r;
}
if(t.max_all<t1.max_suffix+t2.sum || (t.max_all==t1.max_suffix+t2.sum &&t.l>t1.suf_l))
{
t.max_all=t1.max_suffix+t2.sum;
t.l = t1.suf_l;
t.r = t2.r;
}
if(t.max_all<t2.max_all || (t.max_all==t2.max_all&&t.l>t2.l))
{
t.max_all = t2.max_all;
t.l = t2.l;
t.r = t2.r;
}
if(t1.sum+t2.max_prefix>t.max_prefix)
{
t.max_prefix=t1.sum+t2.max_prefix;
t.pre_r = t2.pre_r;
}
if(t2.sum+t1.max_suffix>=t.max_suffix)
{
t.max_suffix=t2.sum+t1.max_suffix;
t.suf_l = t1.suf_l;
}
return t;
}
}
int main()
{
int i,j,l,r,cas = 1;
while(~scanf("%d%d",&n,&m))
{
for(i = 1; i<=n; i++)
scanf("%lld",&s[i]);
init(1,n,1);
printf("Case %d:\n",cas++);
while(m--)
{
scanf("%d%d",&l,&r);
node t=query(1,l,r,1,n);
printf("%d %d\n",t.l,t.r);
}
}
return 0;
}