hdu 4107

Gangster

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3123    Accepted Submission(s): 762

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

3 2 1
1 2 1
2 3 1

 

Sample Output

1 3 1

 

Source

2011 Alibaba-Cup Campus Contest

 

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题意不想说 ，妈的 什么破题目 ，一样的代码第一次交就过，第二次就wa

时间卡，这段时间杭电也卡，我快晕了

线段树，用c++交

 #include<iostream>
 #include<cstdio>
 using namespace std;
 #define N 200004
 int  n,m,p;
 struct node
 {
     int left,right;
     int max,min;
     int num;
 }tree[N*];
 void build(int l,int r,int pos)
 {
     int mid=(l+r)>>;
     tree[pos].left=l;
     tree[pos].right=r;
     tree[pos].max=;
     tree[pos].min=;
     tree[pos].num=;
     if(l==r)
     {
         return ;
     }
     build(l,mid,pos<<);
     build(mid+,r,pos<<|);
 }
 void update(int l,int r,int pos,int v)
 {
     int mid=(tree[pos].left+tree[pos].right)>>;
     if(tree[pos].left==l&&tree[pos].right==r)
     {
         if(tree[pos].min>=p)
         {
             tree[pos].min+=v<<;
             tree[pos].max+=v<<;
             tree[pos].num+=v<<;
             return ;
         }
         if(tree[pos].max<p)
         {
             tree[pos].min+=v;
             tree[pos].max+=v;
             tree[pos].num+=v;
             return ;
         }
     }
     if(tree[pos].num!=)
     {
         tree[pos<<].num+=tree[pos].num;
         tree[pos<<].min+=tree[pos].num;
         tree[pos<<].max+=tree[pos].num;
         tree[pos<<|].num+=tree[pos].num;
         tree[pos<<|].min+=tree[pos].num;
         tree[pos<<|].max+=tree[pos].num;
         tree[pos].num=;
     }
     if(r<=mid)
         update(l,r,pos<<,v);
     else if(l>mid)
         update(l,r,pos<<|,v);
     else
     {
         update(l,mid,pos<<,v);
         update(mid+,r,pos<<|,v);
     }
     if(tree[pos*].max>tree[pos<<|].max)
         tree[pos].max=tree[pos<<].max;
     else
         tree[pos].max=tree[pos<<|].max;
     if(tree[pos*].min>tree[pos<<|].min)
         tree[pos].min=tree[pos<<|].min;
     else
         tree[pos].min=tree[pos<<].min;
 }
 void query(int pos)
 {
     if(tree[pos].left==tree[pos].right)
     {
         if(tree[pos].left!=)
             printf(" ");
         printf("%d",tree[pos].num);
         return;
     }
     if(tree[pos].num!=)
     {
         tree[pos<<].num+=tree[pos].num;
         tree[pos<<|].num+=tree[pos].num;
         tree[pos].num=;
     }
     query(pos<<);
     query(pos<<|);
 }
 int main()
 {
     while(scanf("%d%d%d",&n,&m,&p)>)
     {
         build(,n,);
         while(m--)
         {
             int a,b,c;
             scanf("%d%d%d",&a,&b,&c);
             update(a,b,,c);
         }
         query();
         printf("\n");
     }
     return ;
 }