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  • POJ3468 A Simple Problem with Integers 【段树】+【成段更新】

    A Simple Problem with Integers
    Time Limit: 5000MS   Memory Limit: 131072K
    Total Submissions: 57666   Accepted: 17546
    Case Time Limit: 2000MS

    Description

    You have N integers, A1A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

    Input

    The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
    The second line contains N numbers, the initial values of A1A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
    Each of the next Q lines represents an operation.
    "C a b c" means adding c to each of AaAa+1, ... , Ab. -10000 ≤ c ≤ 10000.
    "Q a b" means querying the sum of AaAa+1, ... , Ab.

    Output

    You need to answer all Q commands in order. One answer in a line.

    Sample Input

    10 5
    1 2 3 4 5 6 7 8 9 10
    Q 4 4
    Q 1 10
    Q 2 4
    C 3 6 3
    Q 2 4
    

    Sample Output

    4
    55
    9
    15

    Hint

    The sums may exceed the range of 32-bit integers.

    2014-9-4 16:25:07更新

    #include <stdio.h>
    #include <string.h>
    #define maxn 100002
    #define lson l, mid, rt << 1
    #define rson mid + 1, r, rt << 1 | 1
    typedef long long ll;
    
    struct Node{
    	ll lazy, sum;
    } tree[maxn << 2];
    
    void pushUp(int rt){
    	tree[rt].sum = tree[rt << 1].sum + tree[rt << 1 | 1].sum;
    }
    
    void pushDown(int l, int r, int rt)
    {
    	int mid = (l + r) >> 1;
    	tree[rt << 1].sum += (mid - l + 1) * tree[rt].lazy;
    	tree[rt << 1 | 1].sum += (r - mid) * tree[rt].lazy;
    	tree[rt << 1].lazy += tree[rt].lazy;
    	tree[rt << 1 | 1].lazy += tree[rt].lazy;
    	tree[rt].lazy = 0;
    }
    
    void build(int l, int r, int rt)
    {
    	tree[rt].lazy = 0;
    	if(l == r){
    		scanf("%lld", &tree[rt].sum);
    		return;
    	}
    	int mid = (l + r) >> 1;
    	build(lson); build(rson);
    	pushUp(rt);
    }
    
    void update(int left, int right, ll val, int l, int r, int rt)
    {
    	if(left == l && right == r){
    		tree[rt].sum += (r - l + 1) * val;
    		tree[rt].lazy += val; return;
    	}
    	int mid = (l + r) >> 1;
    	if(tree[rt].lazy) pushDown(l, r, rt);
    	if(right <= mid) update(left, right, val, lson);
    	else if(left > mid) update(left, right, val, rson);
    	else {
    		update(left, mid, val, lson);
    		update(mid + 1, right, val, rson);
    	}
    	pushUp(rt);
    }
    
    ll query(int left, int right, int l, int r, int rt)
    {
    	if(left == l && right == r) return tree[rt].sum;
    	int mid = (l + r) >> 1;
    	if(tree[rt].lazy) pushDown(l, r, rt);
    	if(right <= mid) return query(left, right, lson);
    	else if(left > mid) return query(left, right, rson);
    	return query(left, mid, lson) + query(mid + 1, right, rson);
    }
    
    int main()
    {
    	int n, m, i, a, b;
    	char com[2];
    	ll c;
    	while(scanf("%d%d", &n, &m) == 2){
    		build(1, n, 1);
    		while(m--){
    			scanf("%s%d%d", com, &a, &b);
    			if(com[0] == 'Q') 
    				printf("%lld
    ", query(a, b, 1, n, 1));
    			else{
    				scanf("%lld", &c);
    				update(a, b, c, 1, n, 1);
    			}
    		}
    	}
    	return 0;
    }


    RE了三次,发现是query函数内的pushdown函数位置放错了。

    改了下,最终A掉了。

    //#define DEBUG
    #include <stdio.h>
    #define maxn 100002
    #define lson l, mid, rt << 1
    #define rson mid + 1, r, rt << 1 | 1
    
    long long tree[maxn << 2], arr[maxn], lazy[maxn << 2];
    
    void pushUp(int rt)
    {
    	tree[rt] = tree[rt << 1] + tree[rt << 1 | 1];
    }
    
    void pushDown(int l, int r, int rt)
    {
    	int mid = (l + r) >> 1;
    	
    	tree[rt << 1] += (mid - l + 1) * lazy[rt];
    	tree[rt << 1 | 1] += (r - mid) * lazy[rt];
    	
    	lazy[rt << 1] += lazy[rt];
    	lazy[rt << 1 | 1] += lazy[rt];
    	lazy[rt] = 0;
    }
    
    void build(int l, int r, int rt)
    {
    	lazy[rt] = 0;
    	if(l == r){
    		tree[rt] = arr[r]; return;
    	}
    	
    	int mid = (l + r) >> 1;
    	build(lson);
    	build(rson);
    	
    	pushUp(rt);
    }
    
    void update(int left, int right, long long val, int l, int r, int rt)
    {
    	if(left == l && right == r){
    		lazy[rt] += val; tree[rt] += val * (r - l + 1); return;
    	} //include l == r
    	
    	if(lazy[rt]) pushDown(l, r, rt);
    	
    	int mid = (l + r) >> 1;
    	if(right <= mid) update(left, right, val, lson);
    	else if(left > mid) update(left, right, val, rson);
    	else{
    		update(left, mid, val, lson);
    		update(mid + 1, right, val, rson);
    	}
    	
    	pushUp(rt);
    }
    
    long long query(int left, int right, int l, int r, int rt)
    {	
    	if(left == l && right == r)
    		return tree[rt];
    		
    	if(lazy[rt]) pushDown(l, r, rt);
    	
    	int mid = (l + r) >> 1;
    	if(right <= mid){
    		return query(left, right, lson);
    	}else if(left > mid){
    		return query(left, right, rson);
    	}else{
    		return query(left, mid, lson) + query(mid + 1, right, rson);
    	}
    }
    
    int main()
    {
    	#ifdef DEBUG
    	freopen("../stdin.txt", "r", stdin);
    	freopen("../stdout.txt", "w", stdout);	
    	#endif
    	
    	int n, q, i, a, b;
    	long long c;
    	char com[2];
    	while(scanf("%d%d", &n, &q) == 2){
    		for(i = 1; i <= n; ++i)
    			scanf("%lld", arr + i);
    		build(1, n, 1);
    		
    		while(q--){
    			scanf("%s%d%d", com, &a, &b);
    			if(com[0] == 'C'){
    				scanf("%lld", &c);
    				update(a, b, c, 1, n, 1);
    			}else printf("%lld
    ", query(a, b, 1, n, 1));
    		}
    	}
    	return 0;
    }


    版权声明:本文博客原创文章,博客,未经同意,不得转载。

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  • 原文地址:https://www.cnblogs.com/zfyouxi/p/4737801.html
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