【HDU】【5376】

digger

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 75    Accepted Submission(s): 27

Problem Description

AFa have n mountains. These mountains in a line。All of mountains have same height initially. Every day , AFa would request ZJiaQ to cut some mountains or put stones in some mountains. And request ZJiaQ report how many mountains belong to “high mountain line”
when all of mountains in a range have same height, and higher than the nearest mountain in left and the nearest mountain in right. the range of mountains called “high mountain line”
of course ,the mountain in the most left and most right can’t be one of “high mountain line”

 

Input

  There are multiply case
In each case, the first line contains 3 integers: n(1<=n<=10^9) , q ( 1 <= q <= 50000), r(0 <= r<= 1000 ).n is the number of mountains. q is the number of days. r is the initial height .
In the next q lines, each line contains 3 integers: l, r, val(1<=l<=r<=n, -1000 <= val <= 1000). Means in the day the mountains’ height in range[l,r] have added by val. but you should let the l,r,val xor ans to get true l,r,val. ans is the answer you printed. 

 

Output

Print q lines, each line contains a single answer.

 

Sample Input

  
  

   
   5 5 0
4 5 87
2 5 -48
3 3 17
4 5 -171
5 5 -494
  
  

 

Sample Output

  
  

   
   0
0
0
1
1
  
  

 

Source

BestCoder Round #50 (div.2)

 

Recommend

hujie   |   We have carefully selected several similar problems for you:   5368  5366  5365  5364  5363 

【L，R)  类似于分布函数，，， 或者是导数。。。 如果 > 0则递增，  < 0 则递减。

然后分类讨论

#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <string>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
using namespace std;

#define X first
#define Y second

map<int,int> m;

int n,q,r;
int ans = 0;
const int INF = 1000000000;
void modify(int x,int v)
{
   if(v == 0) return;
    map<int,int>::iterator it,L,R;
   if(m[x] == 0)
   {
       it = L = R = m.find(x);
       -- L;
       ++ R;
      if(v > 0)
      {
          if (L->Y < 0 && R->Y < 0)    
          	  ans += (R->X -x);
          else if (L->Y < 0 && R->Y > 0) 
              ans = ans;
          else if (L->Y > 0 && R->Y < 0) 
              ans -= (x - L->X);
          else if (L->Y > 0 && R->Y > 0)  
              ans = ans;
      }
      else
      {
          if (L->Y < 0 && R->Y < 0)    
              ans = ans;
          else if (L->Y < 0 && R->Y > 0) 
              ans = ans;
          else if (L->Y > 0 && R->Y < 0)
              ans -= (R->X - x);
          else if (L->Y > 0 && R->Y > 0) 
              ans += (x - L->X);
      }
      m[x] = v;
   }
   else
   {
      it = m.find(x);
      if(it->Y > 0 && it->Y + v > 0 || it->Y < 0 && it->Y + v < 0)
      {
          it->Y += v;
          return;
      }
      L = R = it;
      -- L;
      ++ R;
      int v1 = it->Y;
      int v2 = (it->Y)+ v;
      if((it->Y)> 0)
      {
          if (L->Y < 0 && R->Y < 0)    
              ans -= (R->X - x);
          else if (L->Y < 0 && R->Y > 0) 
              ans = ans;
          else if (L->Y > 0 && R->Y < 0) 
              ans += (x - L->X);
          else if(L->Y > 0 && R->Y > 0)  
              ans = ans;
      }
      else
      {
          if (L->Y < 0 && R->Y < 0)    
              ans = ans;
          else if (L->Y < 0 && R->Y > 0) 
              ans = ans;
          else if (L->Y > 0 && R->Y < 0) 
              ans += (R->X - x);
          else if(L->Y > 0 && R->Y > 0)   
              ans -= (x - L->X);
      }

      m.erase(it);
      modify(x,v2);
        
   }
}
int main()
{
    
    while(scanf("%d%d%d",&n,&q,&r) != EOF)
    {
        ans = 0;
        m.clear();
        m[1] = -INF;
        m[n + 1] = INF;
        for (int i=0;i<q;i++)
        {
            int l,r,v;
            scanf("%d%d%d",&l,&r,&v);

           
            l ^= ans;
            r ^= ans;
            v ^= ans;

   		  if(l > r) swap(l,r);
            modify(l,v);
            modify(r+1,-v);

            printf("%d\n",ans);

        }    
    }

}