HDU - 5775-Bubble Sort（权值线段树）-白红宇

HDU - 5775-Bubble Sort（权值线段树）

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发布时间：2019-06-12

本文共 2682 字，大约阅读时间需要 8 分钟。

P is a permutation of the integers from 1 to N(index starting from 1).

Here is the code of Bubble Sort in C++.

for(int i=1;i<=N;++i)     for(int j=N,t;j>i;—j)         if(P[j-1] > P[j])             t=P[j],P[j]=P[j-1],P[j-1]=t;

After the sort, the array is in increasing order. ?? wants to know the absolute values of difference of rightmost place and leftmost place for every number it reached.

InputThe first line of the input gives the number of test cases T; T test cases follow.

Each consists of one line with one integer N, followed by another line with a permutation of the integers from 1 to N, inclusive.

limits

T <= 20

1 <= N <= 100000

N is larger than 10000 in only one case.

OutputFor each test case output “Case #x: y1 y2 … yN” (without quotes), where x is the test case number (starting from 1), and yi is the difference of rightmost place and leftmost place of number i.Sample Input

233 1 231 2 3

Sample Output

Case #1: 1 1 2Case #2: 0 0 0

Hint

In first case, (3, 1, 2) -> (3, 1, 2) -> (1, 3, 2) -> (1, 2, 3)the leftmost place and rightmost place of 1 is 1 and 2, 2 is 2 and 3, 3 is 1 and 3In second case, the array has already in increasing order. So the answer of every number is 0. 思路：找一下左右的端点 左边取当前位置和最后的位置中最小的 右端点右侧等于比他小的数+当前位置 代码：

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #include
             #include
              
               const int maxn=1e5+5;typedef long long ll;using namespace std;struct node{ int l,r; int num;}tree[maxn<<2];struct node1 { int pos,val; bool friend operator<(node1 x,node1 y) { return x.val
               
                >1; build(m<<1,l,mid); build(m<<1|1,mid+1,r); pushup(m);}void update(int m,int pos,int val){ if(tree[m].l==tree[m].r) { tree[m].num=val; return; } int mid=(tree[m].l+tree[m].r)>>1; if(pos<=mid) { update(m<<1,pos,val); } else { update(m<<1|1,pos,val); } pushup(m);}int query(int m,int l,int r){ if(tree[m].l==l&&tree[m].r==r) { return tree[m].num; } int mid=(tree[m].l+tree[m].r)>>1; if(r<=mid) { return query(m<<1,l,r); } else if(l>mid) { return query(m<<1|1,l,r); } else { return query(m<<1,l,mid)+query(m<<1|1,mid+1,r); } }int ans[maxn];int main(){ int T; cin>>T; int cnt=1; while(T--) { int n; scanf("%d",&n); build(1,0,n); for(int t=1;t<=n;t++) { scanf("%d",&p[t].val); p[t].pos=t; update(1,p[t].val,1); } //sort(p+1,p+n+1); for(int t=1;t<=n;t++) { ans[p[t].val]=abs(min(p[t].val,p[t].pos)-(t+query(1,0,p[t].val-1))); update(1,p[t].val,0); } printf("Case #%d: ",cnt++); for(int t=1;t<=n;t++) { if(t==n) { printf("%d\n",ans[t]); } else { printf("%d ",ans[t]); } } } return 0;}

转载于:https://www.cnblogs.com/Staceyacm/p/11323907.html

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