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http://www.lydsy.com/JudgeOnline/problem.php?id=2120 (题目链接)
题意:给出一个n个数,m个询问,每次询问一个区间或修改一个数,求区间内不同的数有多少个。
solution
分块。每次修改就暴力nlogn重新构块。
对于分块的话,代码很好理解。
代码:
// bzoj2120 #include<algorithm> #include<iostream> #include<cstring> #include<cstdlib> #include<cstdio> #include<cmath> #include<vector> #define MOD 1000000007 #define inf 2147483640 #define LL long long #define free(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout); using namespace std; inline LL getint() { LL x=0,f=1;char ch=getchar(); while (ch>‘9‘ || ch<‘0‘) {if (ch==‘-‘) f=-1;ch=getchar();} while (ch>=‘0‘ && ch<=‘9‘) {x=x*10+ch-‘0‘;ch=getchar();} return x*f; } const int maxn=1000010; int head[maxn],a[maxn],b[maxn],pre[maxn],pos[maxn],n,m,q,block; void reset(int x) { int l=(x-1)*block+1,r=min(x*block,n); for (int i=l;i<=r;i++) pre[i]=b[i]; sort(pre+l,pre+r+1); } void build() { for (int i=1;i<=n;i++) { b[i]=head[a[i]]; head[a[i]]=i; pos[i]=(i-1)/block+1; } for (int i=1;i<=m;i++) reset(i); } int find(int x,int v) { int l=(x-1)*block+1,r=min(x*block,n); int f=l; while (l<=r) { int mid=(l+r)>>1; if (pre[mid]<v) l=mid+1; else r=mid-1; } return l-f; } int query(int l,int r) { int ans=0; if (pos[l]==pos[r]) { for (int i=l;i<=r;i++) if (b[i]<l) ans++; } else { for (int i=l;i<=block*pos[l];i++) if (b[i]<l) ans++; for (int i=block*(pos[r]-1)+1;i<=r;i++) if (b[i]<l) ans++; } for (int i=pos[l]+1;i<pos[r];i++) ans+=find(i,l); return ans; } void modify(int x,int y) { for (int i=1;i<=n;i++) head[a[i]]=0; a[x]=y; for (int i=1;i<=n;i++) { int t=b[i]; b[i]=head[a[i]]; if (t!=b[i]) reset(pos[i]); head[a[i]]=i; } } int main() { scanf("%d%d",&n,&q); for (int i=1;i<=n;i++) scanf("%d",&a[i]); block=int(sqrt(n)); if (n%block>0) m=n/block+1; else m=n/block; build(); while (q--) { char ch[10]; int x,y; scanf("%s%d%d",ch,&x,&y); if (ch[0]==‘Q‘) printf("%d\n",query(x,y)); else modify(x,y); } return 0; }
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原文地址:http://www.cnblogs.com/MashiroSky/p/5914630.html