#6284. 數列分塊入門 8
阿新 • • 發佈:2018-12-10
用flag[i]標記第i塊是否被修改過,每次對a[i]進行操作時,對其進行更新。
更新時將flag[i]重置為-1
//第一行輸入一個數字 n。 // //第二行輸入 n 個數字,第 i 個數字為 ai ,以空格隔開。 // //接下來輸入 n 行詢問,每行輸入三個數字 l r c,以空格隔開。 // //表示先詢問位於[l,r]的數字有多少個是c,再把位於[l,r]的數字都改為c #pragma GCC optimize("Ofast") #pragma comment(linker, "/STACK:102400000,102400000") #pragma GCC target(sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx) #include <vector> #include <iostream> #include <string> #include <map> #include <stack> #include <cstring> #include <queue> #include <list> #include <stdio.h> #include <set> #include <algorithm> #include <cstdlib> #include <cmath> #include <iomanip> #include <cctype> #include <sstream> #include <functional> #include <stdlib.h> #include <time.h> #include <bitset> using namespace std; #define pi acos(-1) #define s_1(x) scanf("%d",&x) #define s_2(x,y) scanf("%d%d",&x,&y) #define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X) #define S_1(x) scan_d(x) #define S_2(x,y) scan_d(x),scan_d(y) #define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z) #define PI acos(-1) #define endl '\n' #define srand() srand(time(0)); #define me(x,y) memset(x,y,sizeof(x)); #define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++) #define close() ios::sync_with_stdio(0); cin.tie(0); #define FOR(x,n,i) for(int i=x;i<=n;i++) #define FOr(x,n,i) for(int i=x;i<n;i++) #define fOR(n,x,i) for(int i=n;i>=x;i--) #define fOr(n,x,i) for(int i=n;i>x;i--) #define W while #define sgn(x) ((x) < 0 ? -1 : (x) > 0) #define bug printf("***********\n"); #define db double #define ll long long #define mp make_pair #define pb push_back typedef long long LL; typedef pair <int, int> ii; const int INF=(1<<31); const LL LINF=0x3f3f3f3f3f3f3f3fLL; const int dx[]={-1,0,1,0,1,-1,-1,1}; const int dy[]={0,1,0,-1,-1,1,-1,1}; const int maxn=1e6+10; const int maxx=1e3+10; const double EPS=1e-8; const double eps=1e-8; const int mod=10007; template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);} template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);} template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));} template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));} template <class T> inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;} while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0'); while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;} inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false; while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;} else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}} if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}} if(IsN) num=-num;return true;} void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');} void print(LL a){ Out(a),puts("");} //freopen( "in.txt" , "r" , stdin ); //freopen( "data.txt" , "w" , stdout ); //cerr << "run time is " << clock() << endl; int n,block,l[maxn],r[maxn],num,belong[maxn]; LL a[maxn],ad[maxn],sum[maxn],mt[maxn],flag[maxn]; //vector<int>seg[2005]; void reset(int x){ if(flag[x]==-1) return ; for(int i=l[x];i<=r[x];i++) a[i]=flag[x]; flag[x]=-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;//塊左端點 塊右端點 r[num]=n; for(int i=1;i<=n;i++){ belong[i]=(i-1)/block+1;//i屬於哪一塊 } me(flag,-1); } int query(int x,int y,int c){ int ans=0; reset(belong[x]); for(int i=x;i<=min(y,r[belong[x]]);i++) (a[i]!=c)?a[i]=c:ans++; if(belong[x]!=belong[y]){ reset(belong[y]); for(int i=l[belong[y]];i<=y;i++) (a[i]!=c)?a[i]=c:ans++; } for(int i=belong[x]+1;i<belong[y];i++){ if(flag[i]!=-1){ (flag[i]==c)?ans+=block:flag[i]=c; } else { for(int j=l[i];j<=r[i];j++) (a[j]!=c)?a[j]=c:ans++; flag[i]=c; } } print(ans); } void solve(){ s_1(n); FOR(1,n,i){ S_1(a[i]); } build(); FOR(1,n,i){ LL x,y,c; S_3(x,y,c); query(x,y,c); } } int main(){ //freopen( "in.txt" , "r" , stdin ); //freopen( "data.txt" , "w" , stdout ); int t=1; //init(); //s_1(t); for(int cas=1;cas<=t;cas++){ //printf("Case #%d: ",cas); solve(); } }