(线段树+并查集) Codeforces Round #416 (Div. 2) E Vladik and Entertaining Flags

In his spare time Vladik estimates beauty of the flags.

Every flag could be represented as the matrix n × m which consists of positive integers.

Let's define the beauty of the flag as number of components in its matrix. We call component a set of cells with same numbers and between any pair of cells from that set there exists a path through adjacent cells from same component. Here is the example of the partitioning some flag matrix into components:

But this time he decided to change something in the process. Now he wants to estimate not the entire flag, but some segment. Segment of flag can be described as a submatrix of the flag matrix with opposite corners at (1, l) and (n, r), where conditions 1 ≤ l ≤ r ≤ m are satisfied.

Help Vladik to calculate the beauty for some segments of the given flag.

Input

First line contains three space-separated integers n, m, q (1 ≤ n ≤ 10, 1 ≤ m, q ≤ 105) — dimensions of flag matrix and number of segments respectively.

Each of next n lines contains m space-separated integers — description of flag matrix. All elements of flag matrix is positive integers not exceeding 106.

Each of next q lines contains two space-separated integers l, r (1 ≤ l ≤ r ≤ m) — borders of segment which beauty Vladik wants to know.

Output

For each segment print the result on the corresponding line.

Example

Input
4 5 4
1 1 1 1 1
1 2 2 3 3
1 1 1 2 5
4 4 5 5 5
1 5
2 5
1 2
4 5
Output
6
7
3
4

Note

Partitioning on components for every segment from first test case:

 

线段树维护每个矩形边缘的情况，合并时利用并查集。

  1 #include <iostream>
  2 #include <string>
  3 #include <algorithm>
  4 #include <cstring>
  5 #include <cstdio>
  6 #include <cmath>
  7 #include <queue>
  8 #include <set>
  9 #include <map>
 10 #include <list>
 11 #include <vector>
 12 #include <stack>
 13 #define mp make_pair
 14 #define MIN(a,b) (a>b?b:a)
 15 #define rank rankk
 16 //#define MAX(a,b) (a>b?a:b)
 17 typedef long long ll;
 18 typedef unsigned long long ull;
 19 const int MAX=1e5+5;
 20 const ll INF=9223372036854775807;
 21 const int N=12;
 22 using namespace std;
 23 const int MOD=1e9+7;
 24 typedef pair<int,int> pii;
 25 const double eps=0.000000001;
 26 int par[MAX];
 27 int a[N][MAX];
 28 int n,m,q;
 29 int id[N<<1];
 30 int find(int x)
 31 {
 32     if(x==par[x])
 33         return x;
 34     return par[x]=find(par[x]);
 35 }
 36 void unite(int x,int y)
 37 {
 38     x=find(x);y=find(y);
 39     par[x]=y;
 40 }
 41 bool same(int x,int y)
 42 {
 43     return find(x)==find(y);
 44 }
 45 struct node
 46 {
 47     int l[N],r[N],sum;//l值在[1,n],r的值在[n+1,2n]
 48 }b[MAX<<2];
 49 void merge(node &lson,node &rson,node &rt,int lright,int rleft)
 50 {
 51     rt.sum=lson.sum+rson.sum;
 52     for(int i=1;i<=n;i++)
 53     {
 54         par[i]=lson.l[i];
 55         par[i+n]=lson.r[i];
 56         par[i+2*n]=rson.l[i]+2*n;
 57         par[i+3*n]=rson.r[i]+2*n;
 58     }
 59     for(int i=1;i<=n;i++)
 60     {
 61         if(a[i][lright]==a[i][rleft]&&!same(i+n,i+2*n))
 62         {
 63             --rt.sum;
 64             unite(i+n,i+2*n);
 65         }
 66     }
 67     for(int i=1;i<=4*n;i++)
 68     {
 69         par[i]=find(par[i]);
 70         id[par[i]]=-1;
 71     }
 72     for(int i=1;i<=n;i++)
 73     {
 74         if(id[par[i]]==-1)
 75             id[par[i]]=i;
 76         rt.l[i]=id[par[i]];
 77     }
 78     for(int i=3*n+1;i<=4*n;i++)
 79     {
 80         if(id[par[i]]==-1)
 81             id[par[i]]=i-2*n;
 82         rt.r[i-3*n]=id[par[i]];
 83     }
 84 }
 85 void build(int l,int r,int k)
 86 {
 87     if(l==r)
 88     {
 89         b[k].sum=0;
 90         for(int i=1;i<=n;i++)
 91         {
 92             if(i==1||a[i-1][l]!=a[i][l])
 93             {
 94                 b[k].l[i]=b[k].r[i]=i;
 95                 ++b[k].sum;
 96             }
 97             else
 98                 b[k].l[i]=b[k].r[i]=b[k].l[i-1];
 99         }
100         return ;
101     }
102     int mid=(l+r)/2;
103     build(l,mid,2*k);
104     build(mid+1,r,2*k+1);
105     merge(b[2*k],b[2*k+1],b[k],mid,mid+1);
106 }
107 void query(int ql,int qr,int l,int r,int k,node &ans)
108 {
109     if(l>=ql&&r<=qr)
110     {
111         ans=b[k];return;
112     }
113     int mid=(l+r)/2;
114     if(qr<=mid)
115         query(ql,qr,l,mid,2*k,ans);
116     else if(ql>mid)
117         query(ql,qr,mid+1,r,2*k+1,ans);
118     else
119     {
120         node p,q;
121         query(ql,qr,l,mid,2*k,p);
122         query(ql,qr,mid+1,r,2*k+1,q);
123         merge(p,q,ans,mid,mid+1);
124     }
125 }
126 int main()
127 {
128     int qs;
129     scanf("%d%d%d",&n,&m,&qs);
130     for(int i=1;i<=n;i++)
131         for(int j=1;j<=m;j++)
132             scanf("%d",&a[i][j]);
133     build(1,m,1);
134     node an;
135     int i;
136     for(i=1;i<=qs;i++)
137     {
138         int ls,rs;
139         scanf("%d%d",&ls,&rs);
140         query(ls,rs,1,m,1,an);
141         printf("%d\n",an.sum);
142     }
143     return 0;
144 }

 

posted @ 2017-05-29 22:35  perplex  阅读(319)  评论(0)  编辑  收藏  举报