ZOJ 3453 Doraemon's Sweet Bullet（线段树区间更新+单点更新+最值询问）

Doraemon likes to shoot his enemies with sweet bullets.

n "enemies" stand in a line, each with a life value li (the index i increases from left to right side and starts from 1). Each time Doraemon shoots a sweet bullet from the right side. The ith bullet has a "critical
range" ki. That is to say, it attacks the first enemy from right side whose life value is equal to or greater than ki. The life value of the attacked enemy will decrease to 1 immediately.

However, Doraemon finds a terrible fact that after an enemy is attacked, it will distribute the sweets to his friends and his friends' life value will increase by 1. For the ith enemy, his f
4000
riends are in a consecutive range from the aith
enemy to the bith. Note that an enemy may be the friend of himself so that after being attacked his life value will be 2.

What's more, if a bullet can't find a target--there doesn't exist an enemy with life greater or equal than ki, all enemies' life will increase by 1.

Now Doraemon wants to know after m bullets were shot, what the maximum life value in all enemies is.

Input
The input contains multiple test cases.

Each test case begins with a line containing a single integer n (1 ≤ n ≤ 100000) indicating the number of enemies.

The following n lines describes enemies, one enemy each line. Each line contains three integers, the initial life value li (1 ≤ li ≤ 10000) and his range of friends ai bi (1 ≤ ai ≤ bi≤
n) - that is, enemies with index from ai to bi(inclusive) are all his friends.

The following line contains an integer m (1 ≤ m ≤ 100000) indicating the number of bullets Doraemon has shot. The last m lines follows, each with a single integer ki(1 ≤ ki ≤ 10000) describing
the bullet's "critical range".

Different cases are separated by a blank line.

Process to the end of input.

Output
For each test case, output the maximum life value after shooting in a single line.

Sample Input

3
3 1 2
4 1 3
5 1 1
3
4
2
1

Sample Output

6

Hint
Doraemon suggests that you should use scanf to read data.

题解：

题意：

多啦B梦有n个敌人，每个敌人有一个v值，每次他从右边开一枪，子弹有一个val值，只要最先大于或者等于val值的那个人v值就会变成1，然后刚开始输入的时候顺带有说这个人的同伙为[l,r]区域的人，这些人知道他中枪了v值就会++，如果子弹没打中任何一个人就会全体v值++，问最后最大的v值是多少

思路：

直接用区间最值更新就行了，就是查询的时候稍微变动一下，本来是不难的一题，结果比赛的时候因为一个pushup写成了pushdown，本来一个小时就应该秒掉的题目检查到了比赛2个小时才发现。。。我为什么那么智障

代码：

#include<algorithm>
#include<iostream>
#include<cstring>
#include<stdio.h>
#include<math.h>
#include<string>
#include<stdio.h>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<deque>
using namespace std;
#define lson k*2
#define rson k*2+1
#define M (t[k].l+t[k].r)/2
#define INF 1008611111
#define ll long long
#define eps 1e-15
struct peo//存人的初始信息
{
int l,r;
int v;
}p[100005];
struct node
{
int l,r;
int tag;//lazy tag，表示要加的值
int maxx;//区间极值
}t[100005*4];
void pushup(int k)//向上传递更新
{
t[k].maxx=max(t[lson].maxx,t[rson].maxx);
}
void Build(int l,int r,int k)//日常建树
{
t[k].l=l;
t[k].r=r;
t[k].tag=0;
if(l==r)
{
t[k].maxx=p[r].v;
return;
}
int mid=M;
Build(l,mid,lson);
Build(mid+1,r,rson);
pushup(k);
}
void pushdown(int k)//日常下推标记
{
if(t[k].tag!=0)
{
t[lson].tag+=t[k].tag;
t[rson].tag+=t[k].tag;
t[lson].maxx+=t[k].tag;
t[rson].maxx+=t[k].tag;
t[k].tag=0;
}
}
void update(int l,int r,int k,int v)//日常区间更新
{
if(l==t[k].l&&t[k].r==r)
{
t[k].tag+=v;
t[k].maxx+=v;
return;
}
pushdown(k);
int mid=M;
if(r<=mid)
update(l,r,lson,v);
else if(l>mid)
update(l,r,rson,v);
else
{
update(l,mid,lson,v);
update(mid+1,r,rson,v);
}
pushup(k);
}
void update1(int pos,int k)//日常单点更新
{
if(t[k].l==pos&&pos==t[k].r)
{
t[k].maxx=1;
return;
}
pushdown(k);
int mid=M;
if(pos<=mid)
update1(pos,lson);
else
update1(pos,rson);
pushup(k);
}
int query(int l,int r,int k,int v)//有点变化的询问
{
if(l==r)
{
return l;
}
pushdown(k);
int mid=M;
if(t[rson].maxx>=v)//如果右边极值满足情况就询问右边
return query(mid+1,r,rson,v);
else
return query(l,mid,lson,v);
}
int main()
{
int i,j,n,m,q,x;
while(scanf("%d",&n)!=EOF)
{
for(i=1;i<=n;i++)
{
scanf("%d%d%d",&p[i].v,&p[i].l,&p[i].r);
}
Build(1,n,1);
scanf("%d",&m);
while(m--)
{
scanf("%d",&x);
if(t[1].maxx<x)//如果最大区间最大值都小于x
{
update(1,n,1,1);
}
else
{
int d=query(1,n,1,x);
update1(d,1);//注意是先单点更新。。一开始我还以为朋友不能包括自己
update(p[d].l,p[d].r,1,1);
}
}
printf("%d\n",t[1].maxx);
}
return 0;
}