HDOJ 题目4107 Gangster（线段树）

最新推荐文章于 2025-06-20 09:03:50 发布

Jogging_Clown

最新推荐文章于 2025-06-20 09:03:50 发布

阅读量470

点赞数

CC 4.0 BY-SA版权

分类专栏：线段树

本文链接：https://blog.youkuaiyun.com/yu_ch_sh/article/details/48728019

线段树专栏收录该内容

61 篇文章

订阅专栏

本文介绍了一个关于两个帮派冲突的问题，其中一个帮派拥有一把能够对特定范围内成员造成伤害的魔法枪。文章通过一个数组来模拟这一过程，并利用线段树进行区间更新和查询操作，以高效地解决此问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Gangster

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2869 Accepted Submission(s): 728

Problem Description

There are two groups of gangsters fighting with each other. The first group stands in a line, but the other group has a magic gun that can shoot a range [a, b], and everyone in that range will take a damage of c points. When a gangster is taking damage, if he has already taken at least P point of damage, then the damage will be doubled. You are required to calculate the damage that each gangster in the first group toke.
To simplify the problem, you are given an array A of length N and a magic number P. Initially, all the elements in this array are 0.
Now, you have to perform a sequence of operation. Each operation is represented as (a, b, c), which means: For each A[i] (a <= i <= b), if A[i] < P, then A[i] will be A[i] + c, else A[i] will be A[i] + c * 2.
Compute all the elements in this array when all the operations finish.

Input

The input consists several testcases.
The first line contains three integers n, m, P (1 <= n, m, P <= 200000), denoting the size of the array, the number of operations and the magic number.
Next m lines represent the operations. Each operation consists of three integers a; b and c (1 <= a <= b <= n, 1 <= c <= 20).

Output

Print A[1] to A[n] in one line. All the numbers are separated by a space.

Sample Input

Sample Output

  
   1 3 1

Source

2011 Alibaba-Cup Campus Contest

Recommend

lcy | We have carefully selected several similar problems for you: 4102 4103 4105 4106 4108

题目大意：输入n,m,p,给你一个有n个数的序列，开始都是0，然后后边m个操作，每个操作

g++过了，c++超时。。弱啊。。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
#define N  200010
int n,m,p;
struct s
{
	int maxn,minn,cover;
}node[N*3];
void build(int l,int r,int tr)
{
	node[tr].maxn=node[tr].minn=node[tr].cover=0;
	int mid=(l+r)>>1;
	if(l==r)
		return;
	build(l,mid,tr<<1);
	build(mid+1,r,tr<<1|1);
}
void pushup(int tr)
{
	node[tr].maxn=max(node[tr<<1].maxn,node[tr<<1|1].maxn);
	node[tr].minn=min(node[tr<<1].minn,node[tr<<1|1].minn);
}
void pushdown(int tr)
{
	if(node[tr].cover)
	{
		node[tr<<1].cover+=node[tr].cover;
		node[tr<<1|1].cover+=node[tr].cover;
		node[tr<<1].maxn+=node[tr].cover;
		node[tr<<1|1].maxn+=node[tr].cover;
		node[tr<<1].minn+=node[tr].cover;
		node[tr<<1|1].minn+=node[tr].cover;
		node[tr].cover=0;
	}
}
void update(int L,int R,int val,int l,int r,int tr)
{
	if(L<=l&&r<=R)
	{
		if(node[tr].maxn<p)
		{
			node[tr].maxn+=val;
			node[tr].minn+=val;
			node[tr].cover+=val;
			return;
		}
		if(node[tr].minn>=p)
		{
			node[tr].minn+=(val<<1);
			node[tr].maxn+=(val<<1);
			node[tr].cover+=(val<<1);
			return;
		}
	}
	pushdown(tr);
	int mid=(l+r)>>1;
	if(L<=mid)
		update(L,R,val,l,mid,tr<<1);
	if(R>mid)
		update(L,R,val,mid+1,r,tr<<1|1);
	pushup(tr);
}
void print(int l,int r,int tr)
{
	if(l==r)
	{
		if(l==1)
			printf("%d",node[tr].maxn);
		else
			printf(" %d",node[tr].maxn);
		return;
	}
	int mid=(l+r)>>1;
	pushdown(tr);
	print(l,mid,tr<<1);
	print(mid+1,r,tr<<1|1);
}
int main()
{
//	int n,m,p;
	while(scanf("%d%d%d",&n,&m,&p)!=EOF)
	{
		build(1,n,1);
		while(m--)
		{
			int a,b,c;
			scanf("%d%d%d",&a,&b,&c);
			update(a,b,c,1,n,1);
		//	print(1,n,1);
		//	printf("\n");
		}
		print(1,n,1);
		printf("\n");
	}
}