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Date Time

16/08/2 23:23

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codeforces

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codeforces 123 D. String　（后缀数组+两次二分得到区间＋rmq）

题目链接

题意：定义一个函数F．．

For exampe: F(babbabbababbab, babb) = 6. The list of pairs is as follows:

(1, 4), (4, 7), (9, 12)

Its continuous sequences are:

  • (1, 4)

  • (4, 7)

  • (9, 12)

  • (1, 4), (4, 7)

  • (4, 7), (9, 12)

  • (1, 4), (4, 7), (9, 12)

    erfen

．
erfen
题目描述得很烂．．看例子把．．大概就是：如果字符串y在字符串x中出现n次，那么F(x,y)=n*(n+1)/2

现在给一个字符串，求所有的F(s,x)的和，x为字符串的所有不相同的子串．

 

思路：由于刚刚写了一个求一个字符串所有不同子串个数的题目．．．于是就想到了后缀数组．．．

写完之后观察height[i]．如果把height[i]看成底在x轴上的第i个矩形的高的话，n就是一段连续的矩形的长度．

 

然后．．．暴力会tle 48

题解说单调栈．．．但是单调栈之后还要线段树or并查集？　（by 羊神）

．．．不会啊orz

最后用了二分+rmp过掉的

大概就是两次二分得到一个矩形的区间，和whust2016 #1的那道题有点像．

C++
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/* ***********************************************
Author :111qqz
Created Time :2016年08月01日 星期一 04时57分01秒
File Name :code/cf/problem/123d.cpp
************************************************ */
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <stack>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair
using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N=1E5+7;
int n,sa[N],rk[N],t[N],t2[N],cnt[N];
int height[N];
char s[N];
int cmp( int *r , int a,int b,int l){return r[a]==r[b]&&r[a+l]==r[b+l];}
void getSa( int n,int m)
{
    int *x = t;
    int *y = t2;
    ms(cnt,0);
    for ( int i = 0 ; i  < n ; i++) cnt[x[i]=s[i]]++;
    for ( int i = 1 ; i < m ; i++) cnt[i] +=cnt[i-1];
    for ( int i = n-1 ; i >= 0 ; i--) sa[--cnt[x[i]]] = i ;
    for ( int k = 1 ; k <= n ; k <<=1)
    {
        int p = 0 ;
        for ( int i = n-k ; i < n;  i++) y[p++] = i;
        for  ( int i = 0 ; i < n ; i++) if (sa[i]>=k) y[p++] = sa[i]-k;
        ms(cnt,0);
        for ( int i = 0 ; i <n ; i++) cnt[x[y[i]]]++;
        for ( int i = 0 ; i < m ; i++) cnt[i] +=cnt[i-1];
        for ( int i = n-1 ; i >= 0 ; i--) sa[--cnt[x[y[i]]]] = y[i];
        swap(x,y);
        p = 1;
        x[sa[0]] = 0 ;
        for ( int i = 1 ;i  <n ; i++)
            x[sa[i]] = cmp(y,sa[i-1],sa[i],k)?p-1:p++;
        if (p>=n) break;
        m = p;
    }
}
void getHeight( int n)
{
    int k = 0 ;
    for ( int i = 0 ; i < n ; i++) rk[sa[i]] = i ;
    height[0] =  0 ;
    for ( int  i = 0 ; i < n ; i++)
    {
        if (rk[i]==0) continue;
        if (k) k--;
        int j = sa[rk[i]-1];
        while (s[i+k]==s[j+k]) k++;
        height[rk[i]] = k;
    }
}
int getSuffix( char s[])
{
    int len =  strlen(s);
    int up = 0 ;
    for ( int i = 0 ; i < len ; i++)
    {
        int  val = s[i];
        up = max(up,val);
    }
    s[len++]='$';
    getSa(len,up+1);
    getHeight(len);
    return len;
}
LL cal( LL x)
{
    return x*(x+1)/2LL;
}
LL solve2(int n)
{
         for ( int i = 0 ; i < n ; i++) cout<<i<<" "<<height[i]<<endl;
    ms(cnt,0);
    int up = 0 ;
    for ( int i = 1 ; i < n;  i++) cnt[height[i]]++,up=max(up,height[i]);
    for ( int i = up-1 ; i >= 0 ; i--) cnt[i] = cnt[i]+cnt[i+1];
    LL res = 0LL ;
    for ( int i = 0 ; i <= up ; i++)
        res = res + cal(1LL*cnt[i]);
    return res;
}
/*
stack<int>stk;
int l[N],r[N];
bool visl[N],visr[N];
 
LL solve ( LL n)  //单调栈
{
    LL res = n*(n-1LL)/2LL;
    int up = 0 ;abaabaabaaba
    for ( int i = 1 ; i < n; i++) up = max(height[i],up);
//   for ( int i = 1 ; i <= n ; i++) cout<<"i:"<<i<<" height[i]:"<<height[i]<<endl;
    height[0] = -1;
    height[n+1] = -1 ; //单调队列的哨兵
    stk.push(0);
    int x;
    for ( int i = 1 ; i <= n; i++)
    {
        for (x=stk.top();height[x]>=height[i];x=stk.top()) stk.pop();
        l[i] = x+1;
        stk.push(i);
//        cout<<"i:"<<i<<endl;
    }
    while (!stk.empty()) stk.pop() ; stk.push(n+1);
    for ( int i  = n ; i >=1 ; i--)
    {
        for (x=stk.top() ; height[x]>=height[i] ; x = stk.top()) stk.pop();
        r[i] = x-1;
        stk.push(i);
    }
  //  for ( int i = 1 ; i <= n ; i++)
//        cout<<"i:"<<i<<" l[i]:"<<l[i]<<" r[i]:"<<r[i]<<" height[i]:"<<height[i]<<" "<<(r[i]-l[i]+1)*height[i]<<endl;
//    int mxh = 0 ;
    ms(visl,false);
    ms(visr,false);
    int mxh = 0 ;
    int more =1;
    for ( int i = 1 ;i  <= n ; i++)
    {
        if (height[i]==0)
        {
            mxh = 0 ;
            continue;
        }
        if (visr[r[i]]&&visl[l[i]])
        {
            mxh = 0 ;
            continue;
        }
            if (i>1&&height[i-1]<height[i]) more = height[i]-height[i-1];
            if (i<n&&height[i+1]<height[i]) more = min(more,height[i]-height[i+1]);
            res = res + cal((r[i]-l[i]+1))*(more);
            mxh = max(mxh,height[i]);
            visr[r[i]] = true;
            visl[l[i]] = true;
//            cout<<"i:"<<i<<" res:"<<res<<endl;
        
      }
        
    return res;
}  */
 
int dp[N][30]; // for rmq;
void rmq_init( int n)
{
    for ( int i = 1 ; i <= n ; i++) dp[i][0] = height[i];
 
    for ( int j = 1 ; (1<<j) <= n ; j++)
        for ( int i = 1 ; i + (1<<j) - 1 <= n ; i++)
            dp[i][j] = min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}
int rmq( int l,int r)
{
    if (l>r) swap(l,r);
    int k = 0 ;
    while (1<<(k+1)<=r-l+1) k++;
    int res = min(dp[l][k],dp[r-(1<<k)+1][k]);
    return res;
 
}
 
LL solve( LL len)
{
    rmq_init(len);
    LL ans = len*(len+1)/2;
 
//   for ( int i = 1 ; i <= len ; i++) cout<<i<<" :"<<height[i]<<endl;
    for ( int i = 1 ; i <= len ; i++)
    {
        int l = 0 ,r = i ;
        while (r>l+1)
        {
            int mid = (l+r)/2;
            if (rmq(mid,i-1)>=height[i]) r = mid;
            else l = mid;
        }
        int L = i,R= len+1;
        while (R>L+1)
        {
            int mid = (L+R) / 2;
            if (rmq(i+1,mid)>height[i]) L = mid;
            else R = mid;
        }
//        cout<<"i:"<<height[i]<<"L:"<<L<<" r:"<<r<<" ans:"<<ans<<endl;
        ans +=1LL*height[i]*(L-i+1)*(i-r+1);
    }
    return ans;
}
int main()
{
        #ifndef  ONLINE_JUDGE
        freopen("code/in.txt","r",stdin);
  #endif
        scanf("%s",s);
        int len = getSuffix(s);
//        cout<<"len:"<<len-1<<endl;
        LL ans = solve(len-1);
        printf("%lld\n",ans);
  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}
 
 
 
 

Date Time

16/03/28 20:37

Author

Category

hdoj

Tags

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bc #77 ||hdu 5652 India and China Origins (图的动态连通性问题，并查集or 二分+bfs验证连通性)

题目链接
题意：没图不好描述，有中文题面中文题面，直接看吧。
思路：据说这道题有三种做法。 当时比赛一种都不会。

先说一种：做法是把格子看成点，可以到达的相邻格子之间看成有边相连，然后倒过来用并查集判断无向图的连通性。具体做法是：先统计初始所有空的位置，然后把所有要增加的山都加上（先统计空的位置是因为山之后要去掉，而去掉以后要得到该点的标号），然后将把所有空的点以及china(设标号为n*m+1)点,和india(设标号为n*m+2) 点通过并查集来合并..可以从上往下从左往右，每次只需要判断上面的点和左边的点是否有空，如果有就用并查集合并。 china点和india点特殊搞就好。

然后判断india和china是否联通，如果是则输出-1.否则从最后添加的山开始移除，每次移除一座山，添加四个方向能添加的边（注意这里不要忘记如果改点在第0行或者第n-1行还要添加和china或者india的边）

然后移除后询问india和china是否联通 （root(china)==root(india)?）

如果时间i联通了，而i+1没有联通，说明时间i是两国最早的失去联系的时间。

 

第一次做这种题目，这种题目的一般做法都是倒过来做。貌似还有一个二分删除的山+bfs判断连通性的。。。？ 窝再搞搞看。
update :二分+bfs判断连通性。其实这个思路更常规。。做法就是字面意思。注意无解的判断即可。

 

并查集解法：

 

 

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/* ***********************************************
Author :111qqz
Created Time :2016年03月27日 星期日 20时11分02秒
File Name :code/bc/#77/1003.cpp
************************************************ */
 
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair
 
using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N=505;
char maze[N][N];
int f[N*N];
int n,m;
int q;
int p[N][N];
int china;
int india;
struct node
{
    int x,y;
    int id;
 
}shan[N*N],kong[N*N];
 
int root ( int x)
{
    if (x!=f[x]) f[x] = root (f[x]);
    return f[x];
}
 
void merge( int x,int y)
{
    int rx = root (x);
    int ry = root (y);
//    cout<<"rx::"<<rx<<"   ry::"<<ry<<endl;
    if (rx!=ry)
    {
        f[rx] = ry ;
//        f[ry] = rx;
    }
}
 
void addegg( int x,int y)
{
    if (x-1>=0&&maze[x-1][y]=='0') merge(p[x][y],p[x-1][y]);
    if (x+1<n&&maze[x+1][y]=='0') merge(p[x][y],p[x+1][y]);
    if (y-1>=0&&maze[x][y-1]=='0') merge(p[x][y],p[x][y-1]);
    if (y+1<m&&maze[x][y+1]=='0') merge(p[x][y],p[x][y+1]);
 
    if (x==0) merge(p[x][y],china);  //一开始忘记更新边缘的山被移除后，点和china以及india的边了。
    if (x==n-1) merge(p[x][y],india);
}
int main()
{
        #ifndef  ONLINE_JUDGE
        freopen("code/in.txt","r",stdin);
  #endif
 
        int T;
        cin>>T;
        while (T--)
        {
            scanf("%d %d",&n,&m);
            for ( int i = 0 ; i < n ; i++) scanf("%s",maze[i]);
 
            scanf("%d",&q);
            for ( int i = 1 ; i <= q ; i++)
            {
                int x,y;
                scanf("%d %d",&x,&y);
                shan[i].x = x;
                shan[i].y = y;
                shan[i].id = i;
//                maze[x][y] = '1';  //先统计空位。
            }
 
            int cnt = 0 ;
            ms(p,-1);
            for ( int i = 0 ; i  < n ; i++)
            {
                for ( int j = 0 ; j < m;  j++)
                {
                    if (maze[i][j]=='0')
                    {
                        cnt++;
                        kong[cnt].x = i ;
                        kong[cnt].y = j;
                        kong[cnt].id = i;
                        p[i][j] = cnt;
                    }
                }
            }
 
            for ( int i  = 1 ; i <= q ; i ++)
            {
                int x = shan[i].x;
                int y = shan[i].y;
                maze[x][y]='1';
            }
            //check kong...ok!
            //for ( int i = 1 ; i  <= cnt ; i++) cout<<"kong:"<<kong[i].x<<" "<<kong[i].y<<endl;
 
                china = n*m+1;
                india = n*m+2;
            for ( int i = 0 ; i <= n*m+2 ; i++) f[i] =  i;
        //    f[china] = china;
          //  f[india] = india;
         //   cout<<"china:"<<china<<endl;
           // cout<<"india:"<<india<<endl;
            for ( int j = 0 ; j  < m ; j++)
            {
                if (maze[0][j]=='0')
                {  
                    merge(china,p[0][j]);
                }
                if (maze[n-1][j]=='0')
                {
                    merge(india,p[n-1][j]);
                }
            }
 
            for ( int i = 0 ;i < n ; i++)
            {
                for ( int j = 0 ; j < m ; j ++)
                {
                    if (maze[i][j]=='1') continue;
                    if (maze[i-1][j]=='0') merge(p[i-1][j],p[i][j]);
                    if (maze[i][j-1]=='0') merge(p[i][j-1],p[i][j]);
                }
            }
 
            if (root(china)==root(india))
            {
                puts("-1");
                continue;
            }
 
            for ( int i = q ; i >= 1 ; i--)
            {
                int x = shan[i].x;
                int y = shan[i].y;
 
                maze[x][y] = '0';
                addegg(x,y);
                if (root(china)==root(india))
                {
                    printf("%d\n",i);
                    break;
                }
            }
        }
 
  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}
 
 
 
 

二分+bfs判断连通性。

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/* ***********************************************
Author :111qqz
Created Time :2016年03月28日 星期一 21时12分05秒
File Name :code/hdu/5652.cpp
************************************************ */
 
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair
 
using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N=501;
char maze[N][N];
int n,m;
int q;
bool vis[N][N];
 
struct node
{
    int x,y;
 
    bool ok ()
    {
        if (x>=0&&x<n&&y>=0&&y<m&&!vis[x][y]) return true;
        return false;
    }
}shan[250005];
 
 
int check( int id)
{
    
    for ( int i = 0 ; i < n ; i++)
    {
        for ( int j = 0 ; j < m ; j ++)
        {
            if (maze[i][j]=='1') vis[i][j] = true;
                else vis[i][j] = false;
        }
    }
 
    for ( int i = 1 ; i <= id ; i++)
    {
        int x = shan[i].x;
        int y = shan[i].y;
        vis[x][y] = true;
    }
 
    queue<node>q;
    for ( int j = 0 ; j  < m ; j++)
    {
        
        if (!vis[0][j])
        {
            node tmp;
            tmp.x = 0;
            tmp.y = j;
            q.push(tmp);
        }
    }
 
    while (!q.empty())
    {
        node pre = q.front();q.pop();
        if (pre.x==n-1) return 1;
        for ( int i = 0 ; i < 4 ; i++)
        {
            node nxt;
            nxt.x = pre.x + dx4[i];
            nxt.y = pre.y + dy4[i];
            if (nxt.ok())
            {
                vis[nxt.x][nxt.y] = true;
                q.push(nxt);
                
            }
        }
    }
    return 0;
}
 
int bin()
{
    int l = 1 ;
    int r = q ;
    int mid;
    while (l<=r)
    {
        mid = (l+r)/2;
        if (check(mid))
            l = mid + 1;
        else r = mid -1;
    }
    if (l>q) return -1;
    return l;
}
int main()
{
        #ifndef  ONLINE_JUDGE
        freopen("code/in.txt","r",stdin);
  #endif
 
        int T;
        cin>>T;
        while (T--)
        {
            scanf("%d%d",&n,&m);
            for ( int i = 0 ; i  < n;  i++) scanf("%s",maze[i]);
            scanf("%d",&q);
            for ( int i = 1 ; i <= q ; i ++) scanf("%d %d",&shan[i].x,&shan[i].y);
            int ans = bin();
            printf("%d\n",ans);
        }
 
  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}
 
 
 
 

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