UVA 12086 Potentiometers（树状数组|| 线段树单点更新）

比眉伴天荒 2022-06-10 02:53 126阅读 0赞

A potentiometer, or potmeter for short, is an electronic device with a variable electric resistance. It has two terminals and some kind of control mechanism (often a dial, a wheel or a slide) with which the resistance between the terminals can be adjusted from zero (no resistance) to some maximum value. Resistance is measured in Ohms, and when two or more resistors are connected in series (one after the other, in a row), the total resistance of the array is the sum of the resistances of the individual resistors.

In this problem we will consider an array of N potmeters, numbered 1 to N from left to right. The left terminal of some potmeter numbered x is connected to the right terminal of potmeter x − 1, and its right terminal to the left terminal of potmeter x + 1. The left terminal of potmeter 1 and the right terminal of potmeter N are not connected.

Initially all the potmeters are set to some value between 0 and 1000 Ohms. Then we can do two things:

• Set one of the potmeters to another value.  
• Measure the resistance between two terminals anywhere in the array.

Input

The input consists less than 3 cases. Each case starts with N, the number of potmeters in the array, on a line by itself. N can be as large as 200000. Each of next N lines contains one numbers between 0 and 1000, the initial resistances of the potmeters in the order 1 to N. Then follow a number of actions, each on a line by itself. The number of actions can be as many as 200000. There are three types of action:

*  “S x r” - set potmeter x to r Ohms. x is a valid potmeter number and r is between 0 and 1000.
 *  “M x y” - measure the resistance between the left terminal of potmeter x and the right terminal
    
    of potmeter y. Both numbers will be valid and x is smaller than or equal to y.
 *  “END” - end of this case. Appears only once at the end of a list of actions.
    
    A case with N = 0 signals the end of the input and it should not be processed. Output
    
    For each case in the input produce a line ‘Case n:’, where n is the case number, starting from 1.  
    For each measurement in the input, output a line containing one number: the measured resistance
    
    in Ohms. The actions should be applied to the array of potmeters in the order given in the input. Print a blank line between cases.
    
    Warning: Input Data is pretty big (∼ 8 MB) so use faster IO. Sample Input
    
    3  
    100  
    100  
    100 M11 M13  
    S 2 200 M12 S30 M23 END  
    10  
    1  
    2  
    3  
    4  
    5  
    6  
    7  
    8  
    9  
    10  
    M 1 10 END  
    0
    
    Sample Output
    
        Case 1:
        100
        300
        300
    
    200
    
    Case 2: 55

题意：

很水的线段树单点更新和区间求和，我直接用树状数组做了

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<deque>
    #define M (t[k].l+t[k].r)/2
    #define lson k*2
    #define rson k*2+1
    #define ll long long
    using namespace std;
    int sum[200005];
    int n;
    int lowbit(int x)
    {
        return x&(-x);
    }
    void update(int x,int v)//日常更新
    {
        while(x<=200001)//这里+1是因为有要-1的操作，放在在查询时x=0无限循环，所以所有边界和值都+1
        {
            sum[x]+=v;
            x+=lowbit(x);
        }
    }
    int query(int x)//日常查询
    {
        int s=0;
        while(x>0)
        {
            s+=sum[x];
            x-=lowbit(x);
        }
        return s;
    }
    int main()
    {
        int i,j,m,x,y;
        int ans=1;
        char s[10];
        while(scanf("%d",&n)!=EOF&&n)
        {
            memset(sum,0,sizeof(sum));
            for(i=1;i<=n;i++)
            {
                scanf("%d",&x);
                update(i+1,x);//防止出问题全部+1
            }
            if(ans!=1)
                printf("\n");
            printf("Case %d:\n",ans);
            ans++;
            while(scanf("%s",s)!=EOF)
            {
                if(s[0]=='M')
                {
                    scanf("%d%d",&x,&y);
                    x++;
                    y++;
                    printf("%d\n",query(y)-query(x-1));//因为这里要-1，所以所有+1，查询[x,y]和相当于[0,y]的和减去[0,x]的和
                }
                else if(s[0]=='S')
                {
                    scanf("%d%d",&x,&y);
                    x++;
                    int t=query(x)-query(x-1);//先减去原来的值，再加上新的值
                    update(x,-t);
                    update(x,y);
                }
                else
                    break;
            }
        }
        return 0;
    }