Gorgeous Sequence
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 349 Accepted Submission(s): 57
Problem Description
There is a sequence a of length n. We use ai to denote the i-th element in this sequence. You should do the following three types of operations to this sequence.
0 x y t: For every x≤i≤y, we use min(ai,t) to replace the original ai's value.
1 x y: Print the maximum value of ai that x≤i≤y.
2 x y: Print the sum of ai that x≤i≤y.
0 x y t: For every x≤i≤y, we use min(ai,t) to replace the original ai's value.
1 x y: Print the maximum value of ai that x≤i≤y.
2 x y: Print the sum of ai that x≤i≤y.
Input
The first line of the input is a single integer T, indicating the number of testcases.
The first line contains two integers n and m denoting the length of the sequence and the number of operations.
The second line contains n separated integers a1,…,an (∀1≤i≤n,0≤ai<231).
Each of the following m lines represents one operation (1≤x≤y≤n,0≤t<231).
It is guaranteed that T=100, ∑n≤1000000, ∑m≤1000000.
Output
For every operation of type 1 or 2, print one line containing the answer to the corresponding query.
Sample Input
1
5 5
1 2 3 4 5
1 1 5
2 1 5
0 3 5 3
1 1 5
2 1 5
Sample Output
5
15
3
12
Hint
Please use efficient IO methodSource
解题：线段树，对着标程写的，还是没搞清到底是怎么回事
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 const int maxn = 1000010; 5 struct node { 6 int cover,s,tag,maxtag,maxv; 7 LL sum; 8 } tree[maxn<<2]; 9 node put(node c,int t) { 10 if(!t) return c; 11 if(c.cover != c.s) c.maxv = t; 12 c.maxtag = t; 13 c.sum += (LL)t*(c.s - c.cover); 14 c.cover = c.s; 15 return c; 16 } 17 node calc(const node &a,const node &b,int t) { 18 node c; 19 c.tag = t; 20 c.s = a.s + b.s; 21 c.sum = a.sum + b.sum; 22 c.cover = a.cover + b.cover; 23 c.maxv = max(a.maxv,b.maxv); 24 c.maxtag = max(a.maxtag,b.maxtag); 25 return put(c,t); 26 } 27 int tmp,n,m; 28 void build(int L,int R,int v) { 29 tree[v].tag = 0; 30 tree[v].s = R - L + 1; 31 if(L == R) { 32 scanf("%d",&tmp); 33 tree[v].sum = tree[v].tag = tree[v].maxtag = tree[v].maxv = tmp; 34 tree[v].cover = 1; 35 return; 36 } 37 int mid = (L + R)>>1; 38 build(L,mid,v<<1); 39 build(mid+1,R,v<<1|1); 40 tree[v] = calc(tree[v<<1],tree[v<<1|1],0); 41 } 42 node query(int L,int R,int lt,int rt,int v) { 43 if(lt <= L && rt >= R) return tree[v]; 44 int mid = (L + R)>>1; 45 if(rt <= mid) return put(query(L,mid,lt,rt,v<<1),tree[v].tag); 46 if(lt > mid) return put(query(mid+1,R,lt,rt,v<<1|1),tree[v].tag); 47 return calc(query(L,mid,lt,rt,v<<1),query(mid+1,R,lt,rt,v<<1|1),tree[v].tag); 48 } 49 void cleartag(int v,int t) { 50 if(tree[v].maxtag < t) return; 51 if(tree[v].tag >= t) tree[v].tag = 0; 52 if(tree[v].s > 1) { 53 cleartag(v<<1,t); 54 cleartag(v<<1|1,t); 55 } 56 if(tree[v].s == 1) { 57 tree[v].sum = tree[v].maxtag = tree[v].maxv = tree[v].tag; 58 tree[v].cover = (tree[v].tag > 0); 59 } else tree[v] = calc(tree[v<<1],tree[v<<1|1],tree[v].tag); 60 } 61 void update(int L,int R,int lt,int rt,int t,int v) { 62 if(tree[v].tag && tree[v].tag <= t) return; 63 if(lt <= L && rt >= R) { 64 cleartag(v,t); 65 tree[v].tag = t; 66 if(L == R) { 67 tree[v].sum = tree[v].tag = tree[v].maxv = tree[v].maxtag = t; 68 tree[v].cover = (tree[v].tag > 0); 69 } else tree[v] = calc(tree[v<<1],tree[v<<1|1],t); 70 } else { 71 int mid = (L + R)>>1; 72 if(rt <= mid) update(L,mid,lt,rt,t,v<<1); 73 else if(lt > mid) update(mid+1,R,lt,rt,t,v<<1|1); 74 else { 75 update(L,mid,lt,rt,t,v<<1); 76 update(mid+1,R,lt,rt,t,v<<1|1); 77 } 78 tree[v] = calc(tree[v<<1],tree[v<<1|1],tree[v].tag); 79 } 80 } 81 int main() { 82 int kase,op,x,y,t; 83 scanf("%d",&kase); 84 while(kase--) { 85 scanf("%d%d",&n,&m); 86 build(1,n,1); 87 while(m--) { 88 scanf("%d%d%d",&op,&x,&y); 89 if(!op) { 90 scanf("%d",&t); 91 update(1,n,x,y,t,1); 92 } else if(op == 1) printf("%d ",query(1,n,x,y,1).maxv); 93 else printf("%I64d ",query(1,n,x,y,1).sum); 94 } 95 } 96 return 0; 97 }
哥终于搞清了，感谢 某岛 的代码
本题的做法有点类似于天龙八部里面虚竹那个下棋场景，死而后生
做法就是：每次更新一段区间的时候，把标记放到本区间，然后，我们要兵马未动，粮草先行
先去下面走一遭，统计下面有多少个不大于t，并且这些不大于t的元素的和
这样就能计算出当前区间的和了？不能？那么最大值呢？别逗了，当前区间的最大值肯定是t啦。
tag?要么就是0，要么就是区间的最大值。不是么？
你说这样都不超时真没天理？是的，我也是这么觉得的。。。每次都推到底，这还不超时。。。我也不知道分摊是如何分摊的
下面是搞清真相后写的另一份代码，代码内附有注释，我想这是你目前所能看到的最详细的解说了，毕竟中学生写的代码，可能是有代沟吧
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 const int maxn = 1001010; 5 struct node{ 6 LL sum; 7 int cv,mx,tag; 8 //cv用来记录该区间有多少个数不大于t 9 //mx当然是记录区间的最大值 10 //tag是lazy标识啦，这个线段树常客 11 //sum... 12 }tree[maxn<<2]; 13 void pushup(int v){ 14 tree[v].sum = tree[v<<1].sum + tree[v<<1|1].sum; 15 tree[v].mx = max(tree[v<<1].mx,tree[v<<1|1].mx); 16 tree[v].cv = tree[v<<1].cv + tree[v<<1|1].cv; 17 } 18 int tmp; 19 void build(int lt,int rt,int v){ 20 if(lt == rt){ 21 scanf("%d",&tmp); 22 tree[v].sum = tree[v].mx = tree[v].tag = tmp; 23 tree[v].cv = 1; 24 //cv记录的是不大于t的元素个数 25 return; 26 } 27 tree[v].tag = 0; 28 int mid = (lt + rt)>>1; 29 build(lt,mid,v<<1); 30 build(mid+1,rt,v<<1|1); 31 pushup(v); 32 } 33 void modify(int L,int R,int t,int v){ 34 if(tree[v].tag && tree[v].tag <= t) return; 35 tree[v].tag = t; 36 //只有可能是tree[v].tag == 0 37 //因为前面calc的时候已经先行走了一遍 38 if(tree[v].cv != R - L + 1){ 39 //因为当前区间不大于t的元素不算全区间，所以还有比t大的 40 //大的要置成t，所以mx成了t 41 tree[v].mx = t; 42 tree[v].sum += (LL)t*(R - L + 1 - tree[v].cv); 43 tree[v].cv = R - L + 1; 44 //sum在alc以后 其实代表的是那些不大于t的元素之和 45 //现在把大于t的都改成t了，那么现在和是多少？ 46 } 47 } 48 void pushdown(int L,int R,int v){ 49 if(tree[v].tag){ 50 int mid = (L + R)>>1; 51 modify(L,mid,tree[v].tag,v<<1); 52 modify(mid+1,R,tree[v].tag,v<<1|1); 53 tree[v].tag = 0; 54 } 55 } 56 void calc(int L,int R,int t,int v){ 57 if(tree[v].mx < t) return; 58 if(tree[v].tag >= t) tree[v].tag = 0; 59 if(L == R){ 60 tree[v].sum = tree[v].mx = tree[v].tag; 61 tree[v].cv = tree[v].tag?1:0; 62 //记录当前不大于t的元素的个数以及他们的和 63 return; 64 } 65 pushdown(L,R,v); 66 int mid = (L + R)>>1; 67 calc(L,mid,t,v<<1); 68 calc(mid+1,R,t,v<<1|1); 69 pushup(v); 70 } 71 void update(int L,int R,int lt,int rt,int t,int v){ 72 if(tree[v].mx <= t) return; 73 if(lt <= L && rt >= R){ 74 if(L == R){ 75 tree[v].sum = tree[v].tag = tree[v].mx = t; 76 tree[v].cv = 1; 77 //强行修改为t,毕竟此处是大于t的，如果不大于t前面return了 78 }else calc(L,R,t,v);//先行遍历到底，兵马未动，粮草先行 79 modify(L,R,t,v); 80 }else{ 81 pushdown(L,R,v); 82 int mid = (L + R)>>1; 83 if(lt <= mid) update(L,mid,lt,rt,t,v<<1); 84 if(rt > mid) update(mid+1,R,lt,rt,t,v<<1|1); 85 pushup(v); 86 } 87 } 88 int queryMax(int L,int R,int lt,int rt,int v){ 89 if(lt <= L && rt >= R) return tree[v].mx; 90 int mid = (L + R)>>1,mx = INT_MIN; 91 pushdown(L,R,v); 92 if(lt <= mid) mx = max(mx,queryMax(L,mid,lt,rt,v<<1)); 93 if(rt > mid) mx = max(mx,queryMax(mid+1,R,lt,rt,v<<1|1)); 94 pushup(v); 95 return mx; 96 } 97 LL querySum(int L,int R,int lt,int rt,int v){ 98 if(lt <= L && rt >= R) return tree[v].sum; 99 int mid = (L + R)>>1; 100 LL sum = 0; 101 pushdown(L,R,v); 102 if(lt <= mid) sum += querySum(L,mid,lt,rt,v<<1); 103 if(rt > mid) sum += querySum(mid+1,R,lt,rt,v<<1|1); 104 pushup(v); 105 return sum; 106 } 107 int main(){ 108 int kase,n,m,op,x,y,t; 109 scanf("%d",&kase); 110 while(kase--){ 111 scanf("%d%d",&n,&m); 112 build(1,n,1); 113 while(m--){ 114 scanf("%d%d%d",&op,&x,&y); 115 if(x > y) swap(x,y); 116 if(!op){ 117 scanf("%d",&t); 118 update(1,n,x,y,t,1); 119 }else if(op == 1) printf("%d ",queryMax(1,n,x,y,1)); 120 else printf("%I64d ",querySum(1,n,x,y,1)); 121 } 122 } 123 return 0; 124 }