BZOJ3236: [Ahoi2013]作业

3236: [Ahoi2013]作业

Time Limit: 100 Sec  Memory Limit: 512 MB
Submit: 702  Solved: 262
[Submit][Status]

Description

Input

Output

Sample Input

3 4
1 2 2
1 2 1 3
1 2 1 1
1 3 1 3
2 3 2 3

Sample Output

2 2
1 1
3 2
2 1

HINT


N=100000,M=1000000

Source

By wangyisong1996加强数据

题解：
明显莫队搞，然后就可以80s'+过掉了。。。
代码：
  1 #include<cstdio>
  2 
  3 #include<cstdlib>
  4 
  5 #include<cmath>
  6 
  7 #include<cstring>
  8 
  9 #include<algorithm>
 10 
 11 #include<iostream>
 12 
 13 #include<vector>
 14 
 15 #include<map>
 16 
 17 #include<set>
 18 
 19 #include<queue>
 20 
 21 #include<string>
 22 
 23 #define inf 1000000000
 24 
 25 #define maxn 1000000+5
 26 
 27 #define maxm 500+100
 28 
 29 #define eps 1e-10
 30 
 31 #define ll long long
 32 
 33 #define pa pair<int,int>
 34 
 35 #define for0(i,n) for(int i=0;i<=(n);i++)
 36 
 37 #define for1(i,n) for(int i=1;i<=(n);i++)
 38 
 39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 40 
 41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 42 
 43 #define mod 1000000007
 44 
 45 using namespace std;
 46 
 47 inline int read()
 48 
 49 {
 50 
 51     int x=0,f=1;char ch=getchar();
 52 
 53     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
 54 
 55     while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
 56 
 57     return x*f;
 58 
 59 }
 60 int n,m,a[maxn],b[maxn],d[maxn],s[2][maxn];
 61 struct rec{int x,y,l,r,id;}c[maxn];
 62 struct recc{int x,y;}ans[maxn];
 63 inline bool cmp(rec x,rec y){return b[x.l]==b[y.l]?x.r<y.r:x.l<y.l;}
 64 inline void add(int z,int x,int y)
 65 {
 66     for(;x<=n;x+=x&(-x))s[z][x]+=y;
 67 }
 68 inline int sum(int z,int x)
 69 {
 70     int t=0;
 71     for(;x;x-=x&(-x))t+=s[z][x];
 72     return t;
 73 }
 74 inline void update(int x,int y)
 75 {
 76     if(y>0)
 77     {
 78         d[x]++;
 79         if(d[x]==1)add(0,x,1);
 80         add(1,x,1);
 81     }
 82     else
 83     {
 84         d[x]--;
 85         if(!d[x])add(0,x,-1);
 86         add(1,x,-1);
 87     }
 88 }
 89 
 90 int main()
 91 
 92 {
 93 
 94     freopen("input.txt","r",stdin);
 95 
 96     freopen("output.txt","w",stdout);
 97 
 98     n=read();m=read();int block=sqrt(n);
 99     for1(i,n)a[i]=read(),b[i]=(i-1)/block+1;
100     for1(i,m)c[i].l=read(),c[i].r=read(),c[i].x=read(),c[i].y=read(),c[i].id=i;
101     sort(c+1,c+m+1,cmp);
102     int l=1,r=0;
103     for1(i,m)
104     {
105         while(r<c[i].r)update(a[++r],1);
106         while(r>c[i].r)update(a[r--],-1);
107         while(l<c[i].l)update(a[l++],-1);
108         while(l>c[i].l)update(a[--l],1);
109         ans[c[i].id].x=sum(1,c[i].y)-sum(1,c[i].x-1);
110         ans[c[i].id].y=sum(0,c[i].y)-sum(0,c[i].x-1);
111     }
112     for1(i,m)printf("%d %d\n",ans[i].x,ans[i].y);
113 
114     return 0;
115 
116 }
View Code

 

posted @ 2014-11-19 16:37  ZYF-ZYF  Views(268)  Comments(0)  Edit  收藏  举报