Light oj 1112 - Curious Robin Hood【线段树-单点更新】题解

野性酷女 2021-07-24 15:35 277阅读 0赞

### 目录 ###

*   *   *  1.题目
         *  2.代码

### 1.题目 ###

Robin Hood likes to loot rich people since he helps the poor people with this money. Instead of keeping all the money together he does another trick. He keeps n sacks where he keeps this money. The sacks are numbered from 0 to n-1.

Now each time he can he can do one of the three tasks.

1.  Give all the money of the ith sack to the poor, leaving the sack empty.
2.  Add new amount (given in input) in the ith sack.
3.  Find the total amount of money from ith sack to jth sack.

Since he is not a programmer, he seeks your help.

Input  
Input starts with an integer T (≤ 5), denoting the number of test cases.

Each case contains two integers n (1 ≤ n ≤ 105) and q (1 ≤ q ≤ 50000). The next line contains n space separated integers in the range \[0, 1000\]. The ith integer denotes the initial amount of money in the ith sack (0 ≤ i < n).

Each of the next q lines contains a task in one of the following form:

1 i Give all the money of the ith (0 ≤ i < n) sack to the poor.

2 i v Add money v (1 ≤ v ≤ 1000) to the ith (0 ≤ i < n) sack.

3 i j Find the total amount of money from ith sack to jth sack (0 ≤ i ≤ j < n).

Output  
For each test case, print the case number first. If the query type is 1, then print the amount of money given to the poor. If the query type is 3, print the total amount from ith to jth sack.

Sample Input

1
    
    5 6
    
    3 2 1 4 5
    
    1 4
    
    2 3 4
    
    3 0 3
    
    1 2
    
    3 0 4
    
    1 1

Sample Output

Case 1:
    
    5
    
    14
    
    1
    
    13
    
    2

### 2.代码 ###

#define _CRT_SECURE_NO_WARNINGS
    #include<iostream>
    #include<cstdio>
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    using namespace std;
    const int maxn = 1e5 + 10;
    int sum[maxn << 2];
    void PushUP(int rt)
    { 
    	sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
    }
    void build(int l, int r, int rt)
    { 
    	if (l == r)
    	{ 
    		scanf("%d", &sum[rt]);
    		return ;
    	}
    	int m = (l + r) >> 1;
    	build(lson);
    	build(rson);
    	PushUP(rt);
    }
    void update(int p, int num, int l, int r, int rt)
    { 
    	if (l == r)
    	{ 
    		if (num < 0)
    		{ 
    			printf("%d\n", sum[rt]);
    			sum[rt] = 0;
    		}
    		else
    		{ 
    			sum[rt] += num;
    
    		}
    		return ;
    	}
    	int m = (l + r) >> 1;
    	if (p <= m)  update(p, num, lson);
    	else         update(p, num, rson);
    	PushUP(rt);
    }
    int query(int L, int R, int l, int r, int rt)
    { 
    	if (L <= l && r <= R)
    	{ 
    		return sum[rt];
    	}
    	int m = (l + r) >> 1;
    	int ans = 0;
    	if (L <= m)  ans += query(L, R,lson);
    	if (R > m)   ans += query(L, R, rson);
    	return ans;
    }
    int main()
    { 
    	int T, n , m;
    	scanf("%d", &T);
    	for (int cas = 1; cas <= T; cas++)
    	{ 
    		scanf("%d%d", &n, &m);
    		build(1, n, 1);
    		printf("Case %d:\n", cas);
    		while(m--)
    		{ 
    			int a, b;
    			int c;
    			scanf("%d", &c);
    			if (c==1)
    			{ 
    				scanf("%d", &a);
    				update(a+1,-1,1,n,1);
    			}
    			else if (c==2)
    			{ 
    				scanf("%d%d", &a, &b);
    				update(a+1, b, 1, n, 1);
    			}
    			else //if(str=='3')
    			{ 
    				scanf("%d%d", &a, &b);
    				printf("%d\n", query(a+1, b+1, 1, n, 1));
    			}
    
    		}
    	}
    	return 0;
    }