BZOJ-3343教主的魔法+分块(大块排序二分)
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传送门:https://www.luogu.org/problemnew/show/P2801
参考:http://hzwer.com/2784.html 感觉思路无比清晰;)
ps:我在洛谷A的,BZOJ要权限;
题意:区间查询有多少个比K的数;
思路:分块,两边暴力更新与查询,中间查询是用二分计数;每次更新,如有必要,要记得重新sort(区间对应的另一个数组);
#include <iostream> #include <cmath> #include <cstdio> #include <cstring> #include <algorithm> #include <string> using namespace std; typedef long long ll; const int maxn = 1000009; const int bmaxn = 1009; ll a[maxn],b[maxn],add[bmaxn]; int n,m; int belong[maxn]; int num,l[bmaxn],r[bmaxn]; //块个数;i这块的左端;i这块的右端 int block; //块大小 void reset(int id) { int le = l[id],ri = r[id]; for(int i=le; i<=ri; i++)b[i] = a[i]; sort(b+le,b+ri+1); } void build() { block = sqrt(n); num = n/block;if(n%block)num++; for(int i=1; i<=num; i++) l[i]=(i-1)*block+1,r[i] = i * block; r[num] = n; for(int i=1; i<=n; i++) belong[i] = (i-1)/block + 1; for(int i=1; i <= num; i++) reset(i); } void update(int lx,int rx,ll val) { if(belong[lx]==belong[rx]) { for(int i=lx;i<=rx;i++)a[i]+=val; reset(belong[lx]); } else { int li = r[belong[lx]],ri = l[belong[rx]]; for(int i = lx; i<=li; i++)a[i]+=val; reset(belong[lx]); for(int i = ri; i<=rx; i++)a[i]+=val; reset(belong[rx]); for(int i = belong[lx] + 1;i < belong[rx]; i++)add[i]+=val; } } ll query(int lx,int rx,ll k) { ll res = 0; if(belong[lx]==belong[rx]) { for(int i=lx;i<=rx;i++)if(a[i] >= k - add[belong[lx]])res++; } else { int li = r[belong[lx]],ri = l[belong[rx]]; // cout<<li<<" "<<ri<<endl; for(int i = lx; i <= li; i++) if(a[i] >= k-add[belong[lx]])res++; for(int i = ri; i <= rx; i++) if(a[i] >= k-add[belong[rx]])res++; for(int i = belong[lx] + 1; i<belong[rx]; i++) { int le = l[i],ri = r[i]; while(le <= ri) { int mid = (le+ri)>>1; if(b[mid] < k - add[i]) le = mid + 1; else ri = mid - 1; } // printf("%d\n",le); // cout<<r[i]-le+1<<endl; res+=r[i] - le + 1; } } return res; } int main(){ scanf("%d%d",&n,&m); for(int i=1; i<=n; i++) scanf("%lld", &a[i]); build(); for(int i=1; i<=m; i++) { char s[20]; int x,y; ll v; scanf("%s%d%d%lld",s,&x,&y,&v); if(s[0]==‘M‘) { update(x,y,v); } else { ll ans = query(x,y,v); printf("%lld\n",ans); } } return 0; }
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