• POJ 3468 A Simple Problem with Integers Splay


    用Splay实现区间增减,查询区间和。

    要对一个区间进行操作只要先把元素在序列中的位置当做键值建树,然后对l,r操作只要把l - 1 splay到根,r+1 splay到根的右子树,那么r + 1的左子树里面就有这个区间的所有的元素了。

    对每个节点存一些信息,就可以很方便的处理了。

    不过这题用splay写相比线段树还是差很多的。。又长又慢。。

    #include <cstdio>
    #include <cstring>
    #include <cstdlib>
    #include <climits>
    
    using namespace std;
    
    const int maxn = 1e6 + 10;
    typedef long long LL;
    int ch[maxn][2], lazy[maxn], fa[maxn], chcnt, root, key[maxn];
    int N, Q, arr[maxn];
    LL sum[maxn], size[maxn], val[maxn];
    
    void pushdown(int x) {
    	if (lazy[x] == 0) return;
    	for (int i = 0; i <= 1; i++) if (ch[x][i]) {
    		int u = ch[x][i];
    		val[u] += lazy[x]; 
    		sum[u] += size[u] * lazy[x];
    		lazy[u] += lazy[x];
    	}
    	lazy[x] = 0;
    }
    
    void pushup(int x) {
    	sum[x] = val[x]; size[x] = 1;
    	for (int i = 0; i <= 1; i++) if (ch[x][i]) {
    		int u = ch[x][i];
    		sum[x] += sum[u]; 
    		size[x] += size[u];
    	}
    }
    
    void rotate(int x, int d) {
    	int y = fa[x];
    	pushdown(y);
    	pushdown(x);
    	fa[ch[x][d]] = y;
    	ch[y][d ^ 1] = ch[x][d];
    	if (fa[y]) ch[fa[y]][y == ch[fa[y]][1]] = x;
    	fa[x] = fa[y]; 
    	fa[y] = x;
    	ch[x][d] = y;
    	pushup(x); pushup(y);
    }
    
    void splay(int x, int goal) {
    	while (fa[x] != goal) {
    		int y = fa[x], d = (x == ch[y][1]);
    		if (fa[y] == goal) rotate(x, d ^ 1);
    		else {
    			int z = fa[y], d1 = (y == ch[z][1]);
    			if (d == d1) {
    				rotate(y, d ^ 1); rotate(x, d ^ 1);
    			}
    			else {
    				rotate(x, d ^ 1); rotate(x, d1 ^ 1);
    			}
    		}
    	}
    	if (goal == 0) root = x;
    }
    
    int access(int v) {
    	int u = root;
    	while (u != 0 && key[u] != v) {
    		pushdown(u);
    		u = ch[u][v > key[u]];
    	}
    	return u;
    }
    
    int accessSeg(int l, int r) {
    	int t = access(l - 1), t1 = access(r + 1);
    	splay(t, 0); 
    	splay(t1, root);
    	return ch[ch[root][1]][0];
    }
    
    void update(int l, int r, int addv) {
    	int u = accessSeg(l, r);
    	lazy[u] += addv; 
    	val[u] += addv; 
    	sum[u] += size[u] * addv;
    }
    
    LL query(int l, int r) {
    	int u = accessSeg(l, r);
    	return sum[u];
    }
    
    int newNode(int &r, int father, int v, int k) {
    	r = ++chcnt; 
    	key[r] = k;
    	fa[r] = father;
    	val[r] = sum[r] = v; 
    	size[r] = 1;
    	lazy[r] = 0; 
    	ch[r][0] = ch[r][1] = 0;
    	return r;
    }
    
    void build(int l, int r, int &rt, int father) {
    	if (l > r) return;
    	int mid = l + r >> 1;
    	newNode(rt, father, arr[mid], mid);
    	if (l == r) return;
    	build(l, mid - 1, ch[rt][0], rt);
    	build(mid + 1, r, ch[rt][1], rt);
    	pushup(rt);
    }
    
    int main() {
    	char cmd;
    	int l, r, cp;
    	while (scanf("%d%d", &N, &Q) != EOF) {
    		chcnt = 0;
    		arr[0] = arr[N + 1] = 0;
    		for (int i = 1; i <= N; i++) {
    			scanf("%d", &arr[i]);
    		}
    		build(0, N + 1, root, 0);
    		for (int i = 1; i <= Q; i++) {
    			scanf(" %c%d%d", &cmd, &l, &r);
    			if (cmd == 'C') {
    				scanf("%d", &cp);
    				update(l, r, cp);
    			}
    			else {
    				printf("%I64d
    ", query(l, r));
    			}
    		}
    	}
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/rolight/p/4277504.html
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