Orz黄学长,蒟蒻在黄学长的带领下,通过阅读黄学长的代码!终于会了这道题! 首先我想先说一下这道题的思路(准确来说是黄学长的)。 很明显,树状数组应该不用讲吧!关键是内存怎么开,维护一些什么样的数据? 其实我们通过观察,很快可以发现,你维护被删的数比维护所有的数轻松多了(不管是空间上,还是时间上)。所以我们就可以从这方面想!(其实我一开始的思路,因为这道题我已经看过很久了,一直想写,毕竟是白书里面的一道例题嘛!一开始,蒟蒻的我是打算这样的用树状数组套权值线段树,并且是维护所有的数,我发现空间不够我开的(或许是我太弱,有大神的话请指教一下。)。好像是因为我维护所有的数才错了,(而且就算要维护,也是动态开点第维护)仔细一想黄学长就是树状数组套权值线段树啊!太弱了我啊!too young too simple 。而且宝宝,开点的时候,一开始只开了一百万,WA掉了,后来看了黄学长开了五百万,自己再仔细一想,logN = 17 , 因为是树套树所以是 logN的平方, 还要乘上操作的次数M, 所以………………,我又错了。加油了多积累经验)。
这是黄学长的树状数组套权值线段树(当然,为了不当简单的做一条源程狗,我自己敲了一遍,并做了修改)
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #define rep(i,j,k) for(int i = j; i <= k; i++) 5 #define down(i,j,k) for(int i = j; i >= k; i--) 6 #define lc c[k][0] 7 #define rc c[k][1] 8 #define N 5000005 9 #define ll long long 10 #define maxn 102333 11 using namespace std; 12 13 int t[maxn], c[N][2], root[maxn], sum[N], a1[maxn], a2[maxn], pos[maxn], num[maxn]; 14 int A[30], B[30]; 15 16 int cnt = 0, n, m; 17 int read() 18 { 19 int s = 0, t = 1; char c = getchar(); 20 while( !isdigit(c) ){ 21 if( c == ‘-‘ ) t = -1; c = getchar(); 22 } 23 while( isdigit(c) ){ 24 s = s * 10 + c - ‘0‘; c = getchar(); 25 } 26 return s * t; 27 } 28 int lower(int x) {return x & (-x); } 29 30 int get_num(int x) 31 { 32 int ans = 0; 33 for( ; x; x -= lower(x)) ans += t[x]; 34 return ans; 35 } 36 37 void update(int&k,int l,int r,int num) 38 { 39 if( !k ) k = ++cnt; 40 sum[k]++; 41 if( l == r ) return; 42 int mid = (l+r)>>1; 43 if( mid < num ) update(rc,mid+1,r,num); 44 else update(lc,l,mid,num); 45 } 46 47 ll get_more(int x,int y,int num) 48 { 49 A[0] = B[0] = 0; x--; ll tmp = 0; 50 for( ; x; x -= lower(x) ) A[++A[0]] = root[x]; 51 for( ; y; y -= lower(y) ) B[++B[0]] = root[y]; 52 int l = 1, r = n; 53 while( l != r ){ 54 int mid = (l+r)>>1; 55 if( num <= mid ) { 56 rep(i,1,A[0]) tmp -= sum[c[A[i]][1]]; 57 rep(i,1,B[0]) tmp += sum[c[B[i]][1]]; 58 rep(i,1,A[0]) A[i] = c[A[i]][0]; 59 rep(i,1,B[0]) B[i] = c[B[i]][0]; 60 r = mid; 61 } 62 else { 63 rep(i,1,A[0]) A[i] = c[A[i]][1]; 64 rep(i,1,B[0]) B[i] = c[B[i]][1]; 65 l = mid+1; 66 } 67 } 68 return tmp; 69 } 70 71 ll get_less(int x,int y,int num) 72 { 73 A[0] = B[0] = 0; x--; ll tmp = 0; 74 for( ; x; x -= lower(x)) A[++A[0]] = root[x]; 75 for( ; y; y -= lower(y)) B[++B[0]] = root[y]; 76 int l = 1, r = n; 77 while( l != r ){ 78 int mid = (l+r)>>1; 79 if( num > mid ){ 80 rep(i,1,A[0]) tmp -= sum[c[A[i]][0]]; 81 rep(i,1,B[0]) tmp += sum[c[B[i]][0]]; 82 rep(i,1,A[0]) A[i] = c[A[i]][1]; 83 rep(i,1,B[0]) B[i] = c[B[i]][1]; 84 l = mid + 1; 85 } 86 else { 87 rep(i,1,A[0]) A[i] = c[A[i]][0]; 88 rep(i,1,B[0]) B[i] = c[B[i]][0]; 89 r = mid; 90 } 91 } 92 return tmp; 93 } 94 95 int main() 96 { 97 n = read(), m = read(); ll ans = 0; 98 rep(i,1,n){ 99 num[i] = read(); pos[num[i]] = i; 100 a1[i] = get_num(n)-get_num(num[i]-1); 101 ans += a1[i]; 102 for(int x = num[i]; x <= n; x += lower(x) ) t[x]++; 103 } 104 memset(t,0,sizeof(t)); 105 down(i,n,1){ 106 int x = num[i]; 107 a2[i] = get_num(x-1); 108 for( ; x <= n; x += lower(x) ) t[x]++; 109 } 110 rep(i,1,m){ 111 printf("%lld\n", ans); 112 int x = read(); int po = pos[x]; 113 ans -= (a1[po] + a2[po] - get_more(1,po-1,x) - get_less(po+1,n,x)); 114 for( ; po <= n; po += lower(po) ) update(root[po],1,n,x); 115 } 116 return 0; 117 }
据翻博客,有人分块A掉了,我不信,怎么可能呢?可是这怎么知道呢?还是还要敲的。很明显,以比树套树的代码慢一倍的代价(大约),A掉了。哈哈!到现在好像还真的没怎么看过分块A不掉的题(不过肯定是有的,是蒟蒻写的题太少,蒟蒻写的题分块几乎都能写),但是写这个不利于自身水平的提高啊!(不到迫不得已,还是少写分块吧!)
1 #include<vector> 2 #include<iostream> 3 #include<cstdio> 4 #include<algorithm> 5 #include<cstring> 6 #define ll long long 7 #define rep(i,j,k) for(int i = j; i <= k; i++) 8 #define down(i,j,k) for(int i = j; i >= k; i--) 9 #define maxn 323 10 #define inf 0x7fffffff 11 using namespace std; 12 13 vector<int> s[maxn]; 14 int pos[100233], v[100233], t[100233]; 15 bool vis[100233]; 16 int n, m; 17 18 int lower(int x ) { return x & (-x); } 19 int getans(int x) 20 { 21 int ans = 0; for(; x; x -= lower(x) ) ans += t[x]; 22 return ans; 23 } 24 25 int read() 26 { 27 int s = 0, t = 1; char c = getchar(); 28 while( !isdigit(c) ){ 29 if( c == ‘-‘ ) t = -1; c = getchar(); 30 } 31 while( isdigit(c) ){ 32 s = s * 10 + c - ‘0‘; c = getchar(); 33 } 34 return s * t; 35 } 36 37 int get_max(int r,int num) 38 { 39 int last = r / maxn; int tmp = 0; 40 rep(i,0,last-1) { 41 tmp += (s[i].size()-1-(upper_bound(s[i].begin(),s[i].end(),num)-s[i].begin())); 42 } 43 int begin = last*maxn; 44 rep(i,begin,r){ 45 if( !vis[i] && v[i] > num ) tmp++; 46 } 47 //cout<<" a "<<tmp<<" "; 48 return tmp; 49 } 50 51 int get_min(int l,int num) 52 { 53 int begin = l / maxn; int tmp = 0; 54 rep(i,begin+1,n/maxn){ 55 tmp += (lower_bound(s[i].begin(),s[i].end(),num)-s[i].begin()); 56 } 57 int last = (begin+1)*maxn-1; 58 rep(i,l,min(last,n)){ 59 // cout<<l+1<<" "<<last<<endl; 60 // cout<<"k "<<i<<endl; 61 if( !vis[i] && v[i] < num ) tmp++; 62 } 63 // cout<<" i "<<tmp<<" "; 64 return tmp; 65 } 66 67 void update(int pos,int num) 68 { 69 int x = pos / maxn; 70 // cout<<"U "<<pos<<endl; 71 s[x].erase(lower_bound(s[x].begin(),s[x].end(),num)); 72 vis[pos] = 1; 73 } 74 75 int main() 76 { 77 n = read(), m = read(); ll ans = 0; 78 rep(i,0,maxn) s[i].push_back(inf); 79 rep(i,1,n){ 80 v[i] = read(); pos[v[i]] = i; s[i/maxn].insert(lower_bound(s[i/maxn].begin(),s[i/maxn].end(),v[i]),v[i]); 81 ans += (getans(n)-getans(v[i]-1)); 82 for(int x = v[i]; x <= n ; x += lower(x) ) t[x]++; 83 } 84 memset(t,0,sizeof(t)); 85 rep(i,1,m){ 86 printf("%lld\n", ans); 87 int x = read(), po = pos[x]; 88 ans -= (get_max(po-1,x)+get_min(po+1,x)); 89 update(po,x); 90 //cout<<endl; 91 } 92 return 0; 93 }
据说此题还有 cdq分治写法,我也orz一下。(这个明天再写了)
时间: 2024-10-15 20:25:22