zoukankan      html  css  js  c++  java

  • 1552: [Cerc2007]robotic sort

    1552: [Cerc2007]robotic sort

    Time Limit: 5 Sec  Memory Limit: 64 MB
    Submit: 1205  Solved: 459

    Description

    Input

    输入共两行，第一行为一个整数N，N表示物品的个数，1<=N<=100000。
    第二行为N个用空格隔开的正整数，表示N个物品最初排列的编号。

    Output

    输出共一行，N个用空格隔开的正整数P1,P2,P3…Pn，Pi表示第i次操作前第i小的物品所在的位置。 
    注意：如果第i次操作前，第i小的物品己经在正确的位置Pi上，我们将区间[Pi,Pi]反转(单个物品)。

    Sample Input

    6
    3 4 5 1 6 2

    Sample Output

    4 6 4 5 6 6

    code

     注意输出格式。 
      1 #include<cstdio>
  2 #include<algorithm>
  3 
  4 using namespace std;
  5 const int MAXN = 100100;
  6 const int INF = 0x7fffffff;
  7 struct DATA{
  8     int x,p;
  9     bool operator < (const DATA &a) const 
 10     {
 11         if (x==a.x) return p < a.p;
 12         return x < a.x; 
 13     }
 14 }d[MAXN];
 15 int mn[MAXN],mnpos[MAXN],data[MAXN],siz[MAXN],fa[MAXN],ch[MAXN][2],tag[MAXN];
 16 int n,root;
 17 
 18 int read()
 19 {
 20     int x = 0,f = 1;char ch = getchar();
 21     while (ch<'0'||ch>'9') {if(ch=='-')f=-1; ch=getchar(); }
 22     while (ch>='0'&&ch<='9') {x=x*10+ch-'0'; ch=getchar();}
 23     return x;
 24 }
 25 void pushup(int x)
 26 {
 27     int l = ch[x][0],r = ch[x][1];
 28     mn[x] = min(data[x],min(mn[l],mn[r]));
 29     if (mn[x]==data[x]) mnpos[x] = x;
 30     else if (mn[x]==mn[l]) mnpos[x] = mnpos[l];
 31     else mnpos[x] = mnpos[r];
 32     siz[x] = siz[l]+siz[r]+1;
 33 }
 34 void pushdown(int x)
 35 {
 36     if (tag[x])
 37     {
 38         tag[x] = 0;
 39         int l = ch[x][0],r = ch[x][1];
 40         tag[l] ^= 1;
 41         swap(ch[l][0],ch[l][1]);
 42         tag[r] ^= 1;
 43         swap(ch[r][0],ch[r][1]);
 44     }
 45 }
 46 int son(int x)
 47 {
 48     return ch[fa[x]][1]==x;
 49 }
 50 void rotate(int x)
 51 {
 52     int y = fa[x],z = fa[y],b = son(x),c = son(y),a = ch[x][!b];
 53     if (z) ch[z][c] = x;else root = x;fa[x] = z;
 54     if (a) fa[a] = y;ch[y][b] = a;
 55     ch[x][!b] = y;fa[y] = x;
 56     pushup(y);pushup(x);
 57 }
 58 void splay(int &x,int rt)
 59 {
 60     while (fa[x]!=rt)
 61     {
 62         int y = fa[x],z = fa[y];
 63         if (z==rt) rotate(x);
 64         else
 65         {
 66             pushdown(z);
 67             pushdown(y);
 68             pushdown(x);
 69             if (son(x)==son(y)) rotate(y),rotate(x);
 70             else rotate(x), rotate(x);
 71         }
 72     }
 73 }
 74 int getkth(int rt,int k)
 75 {
 76     pushdown(rt);
 77     int l = ch[rt][0],r = ch[rt][1];
 78     if (k==siz[l]+1) return rt;
 79     else if (k<siz[l]+1) return getkth(l,k);
 80     else return getkth(r,k-siz[l]-1);
 81 }
 82 int getmnpos(int l,int r)
 83 {
 84     int tl = getkth(root,l-1);
 85     int tr = getkth(root,r+1);
 86     splay(tl,0);
 87     splay(tr,tl);
 88     return mnpos[ch[tr][0]];
 89 }
 90 void reverse(int l,int r)
 91 {
 92     int tl = getkth(root,l-1);//根节点是root 
 93     int tr = getkth(root,r+1);
 94     splay(tl,0);
 95     splay(tr,tl);
 96     int p = ch[tr][0];
 97     tag[p] ^= 1;
 98     swap(ch[p][0],ch[p][1]);
 99 }
100 int main()
101 {
102     n = read();
103     for (int i=2; i<=n+1; ++i)
104     {
105         data[i] = read();
106         d[i].x = data[i];
107         d[i].p = i;
108     }
109     sort(d+2,d+n+2);
110     for (int i=2; i<=n+1; ++i)
111         data[d[i].p] = i;
112     for (int i=0; i<=n+2; ++i)
113         mn[i] = INF;
114     data[0] = data[1] = data[n+2] = INF;
115     siz[0] = 0;root = 1;
116     for (int i=1; i<=n+2; ++i)
117     {
118         fa[i] = i-1;
119         ch[i][1] = i+1;
120     }
121     ch[n+2][1] = 0;
122     for (int i=1; i<=n+2; ++i)
123         pushup(i);
124     for (int i=1; i<=n; ++i)
125     {
126         int p = getmnpos(i+1,n+1);
127         splay(p,0);
128         printf("%d",siz[ch[p][0]]);
129         reverse(i+1,siz[ch[p][0]]+1);
130         if (i!=n) printf(" ");
131     }
132     return 0;
133 }

     标记下传改了下写法

      1 #include<cstdio>
  2 #include<algorithm>
  3 
  4 using namespace std;
  5 const int MAXN = 100100;
  6 const int INF = 0x7fffffff;
  7 struct DATA{
  8     int x,p;
  9     bool operator < (const DATA &a) const 
 10     {
 11         if (x==a.x) return p < a.p;
 12         return x < a.x; 
 13     }
 14 }d[MAXN];
 15 int mn[MAXN],mnpos[MAXN],data[MAXN],siz[MAXN],fa[MAXN],ch[MAXN][2],tag[MAXN];
 16 int n,root;
 17 
 18 int read()
 19 {
 20     int x = 0,f = 1;char ch = getchar();
 21     while (ch<'0'||ch>'9') {if(ch=='-')f=-1; ch=getchar(); }
 22     while (ch>='0'&&ch<='9') {x=x*10+ch-'0'; ch=getchar();}
 23     return x;
 24 }
 25 void pushup(int x)
 26 {
 27     int l = ch[x][0],r = ch[x][1];
 28     mn[x] = min(data[x],min(mn[l],mn[r]));
 29     if (mn[x]==data[x]) mnpos[x] = x;
 30     else if (mn[x]==mn[l]) mnpos[x] = mnpos[l];
 31     else mnpos[x] = mnpos[r];
 32     siz[x] = siz[l]+siz[r]+1;
 33 }
 34 void pushdown(int x)
 35 {
 36     if (tag[x])
 37     {
 38         int l = ch[x][0],r = ch[x][1];
 39         tag[x] = 0;
 40         tag[l] ^= 1;tag[r] ^= 1;
 41         swap(ch[x][1],ch[x][0]);
 42     }
 43 }
 44 int son(int x)
 45 {
 46     return ch[fa[x]][1]==x;
 47 }
 48 void rotate(int x)
 49 {
 50     int y = fa[x],z = fa[y],b = son(x),c = son(y),a = ch[x][!b];
 51     if (z) ch[z][c] = x;else root = x;fa[x] = z;
 52     if (a) fa[a] = y;ch[y][b] = a;
 53     ch[x][!b] = y;fa[y] = x;
 54     pushup(y);pushup(x);
 55 }
 56 void splay(int &x,int rt)
 57 {
 58     while (fa[x]!=rt)
 59     {
 60         int y = fa[x],z = fa[y];
 61         if (z==rt) rotate(x);
 62         else
 63         {
 64             pushdown(z);
 65             pushdown(y);
 66             pushdown(x);
 67             if (son(x)==son(y)) rotate(y),rotate(x);
 68             else rotate(x), rotate(x);
 69         }
 70     }
 71 }
 72 int getkth(int rt,int k)
 73 {
 74     pushdown(rt);
 75     int l = ch[rt][0],r = ch[rt][1];
 76     if (k==siz[l]+1) return rt;
 77     else if (k<siz[l]+1) return getkth(l,k);
 78     else return getkth(r,k-siz[l]-1);
 79 }
 80 int getmnpos(int l,int r)
 81 {
 82     int tl = getkth(root,l-1);
 83     int tr = getkth(root,r+1);
 84     splay(tl,0);
 85     splay(tr,tl);
 86     return mnpos[ch[tr][0]];
 87 }
 88 void reverse(int l,int r)
 89 {
 90     int tl = getkth(root,l-1);//根节点是root 
 91     int tr = getkth(root,r+1);
 92     splay(tl,0);
 93     splay(tr,tl);
 94     tag[ch[tr][0]] ^= 1;
 95 }
 96 int main()
 97 {
 98     n = read();
 99     for (int i=2; i<=n+1; ++i)
100     {
101         data[i] = read();
102         d[i].x = data[i];
103         d[i].p = i;
104     }
105     sort(d+2,d+n+2);
106     for (int i=2; i<=n+1; ++i)
107         data[d[i].p] = i;
108     for (int i=0; i<=n+2; ++i)
109         mn[i] = INF;
110     data[0] = data[1] = data[n+2] = INF;
111     siz[0] = 0;root = 1;
112     for (int i=1; i<=n+2; ++i)
113     {
114         fa[i] = i-1;
115         ch[i][1] = i+1;
116     }
117     ch[n+2][1] = 0;
118     for (int i=1; i<=n+2; ++i)
119         pushup(i);
120     for (int i=1; i<=n; ++i)
121     {
122         int p = getmnpos(i+1,n+1);
123         splay(p,0);
124         printf("%d",siz[ch[p][0]]);
125         reverse(i+1,siz[ch[p][0]]+1);
126         if (i!=n) printf(" ");
127     }
128     return 0;
129 }
    View Code
     
  • 相关阅读:
    Java vs Python
    Compiled Language vs Scripting Language
    445. Add Two Numbers II
    213. House Robber II
    198. House Robber
    276. Paint Fence
    77. Combinations
    54. Spiral Matrix
    82. Remove Duplicates from Sorted List II
    80. Remove Duplicates from Sorted Array II
  • 原文地址：https://www.cnblogs.com/mjtcn/p/7349671.html
Copyright © 2011-2022 走看看