1005 琐碎的区间（线段树）

1005: 琐碎的区间

时间限制: 4 Sec  内存限制: 256 MB
提交: 100  解决: 10
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题目描述

给出一个长度为 n 的整数序列 A[1..n]，有三种操作： 
1 l r x :  把[l, r]区间的每个数都加上 x 
2 l r :  把[l, r]  区间每个 A[i]变为sqrt(a[i])的整数部分
3 l r :  求[l, r]  区间所有数的和 
其中 l 和 r 和 x 都代表一个整数 

输入

第一行一个 T，表示数据组数。 
对于每组数据 
Line1：两个数 n m，表示整数序列长度和操作数 
Line2：n 个数，表示 A[1..n] 
Line3…Line3+m-1：每行一个询问，对于第三种询问，请输出答案。 
对于每一种询问，先给出操作的编号，再给出相应的操作，编号与题目描述对应。 
数据约定： 
1<=T<=5 
n,m <= 100000 
1<= A[i], x<=100000 

输出

对于第三种询问，输出答案。每个答案占一行。 

样例输入

1
5 5
1 2 3 4 5
1 3 5 2
2 1 4
3 2 4
2 3 5
3 1 5

样例输出

5
6

提示

区间维护就好了，关于开方，开个几次就会不变了，此时就可以区间更新了。

首先，这个题，我的个乖乖，超时...这个题还有什么优化吗？

照着网上的搞了一天，还是超时，日后再研究吧，

线段树用数组会比结构体快吗...

先上一个网上的代码吧，

  1 #include <bits/stdc++.h>
  2    
  3 #define ll long long
  4 const int maxn=1e5+10;
  5 const int N=2e5+10;
  6 using namespace std;
  7 #define ls rt<<1
  8 #define rs rt<<1|1
  9 int n,m,k,t,a[maxn];
 10 ll tag[maxn<<2],ma[maxn<<2],mi[maxn<<2],sum[maxn<<2];
 11 const int BufferSize=1<<16;
 12 char buffer[BufferSize],*head,*tail;
 13 inline char Getchar() {
 14     if(head==tail) {
 15         int l=fread(buffer,1,BufferSize,stdin);
 16         tail=(head=buffer)+l;
 17     }
 18     return *head++;
 19 }
 20 inline int read() {
 21     int x=0,f=1;char c=Getchar();
 22     for(;!isdigit(c);c=Getchar()) if(c=='-') f=-1;
 23     for(;isdigit(c);c=Getchar()) x=x*10+c-'0';
 24     return x*f;
 25 }
 26 void pup(int l,int r,int rt)
 27 {
 28     int mid=l+r>>1;
 29     sum[rt]=sum[ls]+sum[rs];
 30     ma[rt]=max(ma[ls],ma[rs]);
 31     mi[rt]=min(mi[ls],mi[rs]);
 32     tag[rt]=0;
 33 }
 34 void pdw(int l,int r,int rt)
 35 {
 36     int mid=l+r>>1;
 37     sum[ls]+=tag[rt]*(mid-l+1);
 38     ma[ls]+=tag[rt];
 39     mi[ls]+=tag[rt];
 40     tag[ls]+=tag[rt];
 41     sum[rs]+=tag[rt]*(r-mid);
 42     ma[rs]+=tag[rt];
 43     mi[rs]+=tag[rt];
 44     tag[rs]+=tag[rt];
 45     tag[rt]=0;
 46 }
 47 void build(int l,int r,int rt)
 48 {
 49     if(l==r)
 50     {
 51         tag[rt]=0;
 52         ma[rt]=mi[rt]=sum[rt]=a[l];
 53         return;
 54     }
 55     int mid=l+r>>1;
 56     build(l,mid,ls);
 57     build(mid+1,r,rs);
 58     pup(l,r,rt);
 59 }
 60 void upd(int L,int R,ll v,int l,int r,int rt)
 61 {
 62     if(L<=l&&r<=R)
 63     {
 64         sum[rt]+=v*(r-l+1);
 65         ma[rt]+=v;
 66         mi[rt]+=v;
 67         tag[rt]+=v;
 68         return;
 69     }
 70     int mid=l+r>>1;
 71     if(tag[rt])pdw(l,r,rt);
 72     if(L<=mid)upd(L,R,v,l,mid,ls);
 73     if(R>mid)upd(L,R,v,mid+1,r,rs);
 74     pup(l,r,rt);
 75 }
 76 void qsqrt(int L,int R,int l,int r,int rt)
 77 {
 78     if(L<=l&&r<=R)
 79     {
 80         if(ma[rt]==mi[rt])
 81         {
 82             tag[rt]-=ma[rt];
 83             ma[rt]=sqrt(ma[rt]);
 84             tag[rt]+=ma[rt];
 85             mi[rt]=ma[rt];
 86             sum[rt]=(r-l+1)*ma[rt];
 87             return;
 88         }
 89         else if(ma[rt]==mi[rt]+1)
 90         {
 91             if((ll)sqrt(ma[rt])==(ll)sqrt(mi[rt])+1)
 92             {
 93                 tag[rt]-=ma[rt];
 94                 sum[rt]-=(r-l+1)*(ma[rt]-(ll)sqrt(ma[rt]));
 95                 ma[rt]=sqrt(ma[rt]);
 96                 tag[rt]+=ma[rt];
 97                 mi[rt]=ma[rt]-1;
 98                 return;
 99             }
100         }
101     }
102     int mid=l+r>>1;
103     if(tag[rt])pdw(l,r,rt);
104     if(L<=mid)qsqrt(L,R,l,mid,ls);
105     if(R>mid)qsqrt(L,R,mid+1,r,rs);
106     pup(l,r,rt);
107 }
108 ll gao(int L,int R,int l,int r,int rt)
109 {
110     if(L<=l&&r<=R)return sum[rt];
111     int mid=l+r>>1;
112     if(tag[rt])pdw(l,r,rt);
113     ll ret=0;
114     if(L<=mid)ret+=gao(L,R,l,mid,ls);
115     if(R>mid)ret+=gao(L,R,mid+1,r,rs);
116     return ret;
117 }
118 int main()
119 {
120     t=read();
121     while(t--)
122     {
123         n=read();m=read();
124         for(int i=1;i<=n;i++)
125             a[i]=read();
126         build(1,n,1);
127         while(m--)
128         {
129             int b,c,d,e;
130             b=read(),c=read(),d=read();
131             if(b==1)
132             {
133                 e=read();
134                 upd(c,d,e,1,n,1);
135             }
136             else if(b==2)
137             {
138                 qsqrt(c,d,1,n,1);
139             }
140             else printf("%lld
",gao(c,d,1,n,1));
141         }
142     }
143     return 0;
144 }
View Code

超时代码，

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 
  4 const int BufferSize=1<<16;
  5 char buffer[BufferSize],*head,*tail;
  6 inline char Getchar() {
  7     if(head==tail) {
  8         int l=fread(buffer,1,BufferSize,stdin);
  9         tail=(head=buffer)+l;
 10     }
 11     return *head++;
 12 }
 13 
 14 inline int read() {
 15     int x=0,f=1;char c=Getchar();
 16     for(;!isdigit(c);c=Getchar()) if(c=='-') f=-1;
 17     for(;isdigit(c);c=Getchar()) x=x*10+c-'0';
 18     return x*f;
 19 }
 20 
 21 #define LL long long
 22 
 23 #define L(root) ((root) << 1)
 24 #define R(root) (((root) << 1) | 1)
 25 
 26 const int MAXN = 1e5 + 5;
 27 int numbers[MAXN];
 28 
 29 //LL delay[MAXN * 4], sum[MAXN * 4], mx[MAXN * 4], mn[MAXN * 4];
 30 
 31 struct Node {
 32     int left, right;
 33     LL delay;
 34     LL sum;
 35     LL mx, mn;
 36     int mid()
 37     {
 38         return left + ((right - left) >> 1);
 39     }
 40 } tree[MAXN * 4];
 41 
 42 void pushUp(int root)
 43 {
 44     tree[root].sum = tree[L(root)].sum + tree[R(root)].sum;
 45     tree[root].mx = max(tree[L(root)].mx, tree[R(root)].mx);
 46     tree[root].mn = min(tree[L(root)].mn, tree[R(root)].mn);
 47     tree[root].delay = 0;
 48 }
 49 
 50 void pushDown(int root, int l, int r)
 51 {
 52         LL mid = (r + l) >> 1;
 53         tree[L(root)].delay += tree[root].delay;
 54         tree[R(root)].delay += tree[root].delay;
 55         tree[L(root)].sum += tree[root].delay * (mid - l + 1);
 56         tree[R(root)].sum += tree[root].delay * (r - mid);
 57         tree[L(root)].mx += tree[root].delay;
 58         tree[R(root)].mx += tree[root].delay;
 59         tree[L(root)].mn += tree[root].delay;
 60         tree[R(root)].mn += tree[root].delay;
 61         tree[root].delay = 0;
 62 
 63 }
 64 
 65 void build(int root, int left, int right)
 66 {
 67     tree[root].left = left;
 68     tree[root].right = right;
 69     if (left == right) {
 70         tree[root].delay = 0;
 71         tree[root].sum = numbers[left];
 72         tree[root].mx = numbers[left];
 73         tree[root].mn = numbers[left];
 74         return;
 75     }
 76     int mid = tree[root].mid();
 77     build(L(root), left, mid);
 78     build(R(root), mid + 1, right);
 79     pushUp(root);
 80 }
 81 
 82 LL query(int root, int left, int right)
 83 {
 84     if (tree[root].left == left && tree[root].right == right) {
 85         return tree[root].sum;
 86     }
 87     if (tree[root].delay) pushDown(root, tree[root].left, tree[root].right);
 88     int mid = tree[root].mid();
 89     if (right <= mid) {
 90         return query(L(root), left, right);
 91     } else if (left > mid) {
 92         return query(R(root), left, right);
 93     } else {
 94         return query(L(root), left, mid) + query(R(root), mid + 1, right);
 95     }
 96 }
 97 
 98 void update(int root, int left, int right, LL add)
 99 {
100     if (tree[root].left == left && tree[root].right == right) {
101         tree[root].delay += add;
102         tree[root].sum += add * (right - left + 1);
103         tree[root].mx += add;
104         tree[root].mn += add;
105         return;
106     }
107     if (tree[root].delay) pushDown(root, tree[root].left, tree[root].right);
108     int mid = tree[root].mid();
109     if (right <= mid) {
110         update(L(root), left, right, add);
111     } else if (left > mid) {
112         update(R(root), left, right, add);
113     } else {
114         update(L(root), left, mid, add);
115         update(R(root), mid + 1, right, add);
116     }
117     pushUp(root);
118 }
119 
120 void sq(int root, int left, int right)
121 {
122     if (tree[root].left == left && tree[root].right == right) {
123         LL mx = sqrt(tree[root].mx);
124         LL mn = sqrt(tree[root].mn);
125         if (tree[root].mx == tree[root].mn) {
126             tree[root].delay -= (tree[root].mx - mx);
127             tree[root].sum = mx * (right - left + 1);
128             tree[root].mx = mx;
129             tree[root].mn = mn;
130             return;
131 //        }
132         } else if ((tree[root].mx == tree[root].mn + 1) && (mx == mn + 1)) {
133             tree[root].delay -= (tree[root].mx - mx);
134             tree[root].sum -= (tree[root].mx - mx) * (right - left + 1);
135             tree[root].mx = mx;
136             tree[root].mn = mn;
137         }
138     }
139     if (tree[root].delay) pushDown(root, tree[root].left, tree[root].right);
140     int mid = tree[root].mid();
141     if (right <= mid) {
142         sq(L(root), left, right);
143     } else if (left > mid) {
144         sq(R(root), left, right);
145     } else {
146         sq(L(root), left, mid);
147         sq(R(root), mid + 1, right);
148     }
149     pushUp(root);
150 }
151 
152 int main()
153 {
154     int t;
155     int n, m;
156     int i;
157     int op, l, r, x;
158 
159 //    scanf("%d", &t);
160     t = read();
161     while (t--) {
162 //        scanf("%d%d", &n, &m);
163         n = read(), m = read();
164         for (i = 1; i <= n; ++i) {
165 //            scanf("%d", &numbers[i]);
166             numbers[i] = read();
167         }
168         build(1, 1, n);
169         for (i = 0; i < m; ++i) {
170 //            scanf("%d", &op);
171             op = read();
172             if (op == 1) {
173 //                scanf("%d%d%d", &l, &r, &x);
174                 l = read(), r = read(), x = read();
175                 update(1, l, r, x);
176                 //printf("debug op = 1
");
177             } else if (op == 2) {
178 //                scanf("%d%d", &l, &r);
179                 l = read(), r = read();
180                 sq(1, l, r);
181                 //printf("debug op = 2
");
182             } else {
183 //                scanf("%d%d", &l, &r);
184                 l = read(), r = read();
185                 printf("%lld
", query(1, l, r));
186                 //printf("debug op = 3
");
187             }
188         }
189     }
190     return 0;
191 }
192 
193 
194 int main2()
195 {
196     int t;
197     int n, m;
198     int i;
199     int op, l, r, x;
200 
201     scanf("%d", &t);
202 //    t = read();
203     while (t--) {
204         scanf("%d%d", &n, &m);
205 //        n = read(), m = read();
206         for (i = 1; i <= n; ++i) {
207             scanf("%d", &numbers[i]);
208 //            numbers[i] = read();
209         }
210         build(1, 1, n);
211         for (i = 0; i < m; ++i) {
212             scanf("%d", &op);
213 //            op = read();
214             if (op == 1) {
215                 scanf("%d%d%d", &l, &r, &x);
216 //                l = read(), r = read(), x = read();
217                 update(1, l, r, x);
218                 //printf("debug op = 1
");
219             } else if (op == 2) {
220                 scanf("%d%d", &l, &r);
221 //                l = read(), r = read();
222                 sq(1, l, r);
223                 //printf("debug op = 2
");
224             } else {
225                 scanf("%d%d", &l, &r);
226 //                l = read(), r = read();
227                 printf("%lld
", query(1, l, r));
228                 //printf("debug op = 3
");
229             }
230         }
231     }
232     return 0;
233 }
View Code
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原文地址：https://www.cnblogs.com/gongpixin/p/6790558.html