PAT-2019年冬季考试-甲级 7-2 Block Reversing (25分) (链表转置)

7-2 Block Reversing (25分)

 

Given a singly linked list L. Let us consider every K nodes as a block (if there are less than K nodes at the end of the list, the rest of the nodes are still considered as a block). Your job is to reverse all the blocks in L. For example, given L as 1→2→3→4→5→6→7→8 and K as 3, your output must be 7→8→4→5→6→1→2→3.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (≤) which is the total number of nodes, and a positive K (≤) which is the size of a block. The address of a node is a 5-digit nonnegative integer, and NULL is represented by −.

Then N lines follow, each describes a node in the format:

Address Data Next

where Address is the position of the node, Data is an integer, and Next is the position of the next node.

Output Specification:

For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 8 3
71120 7 88666
00000 4 99999
00100 1 12309
68237 6 71120
33218 3 00000
99999 5 68237
88666 8 -1
12309 2 33218

Sample Output:

71120 7 88666
88666 8 00000
00000 4 99999
99999 5 68237
68237 6 00100
00100 1 12309
12309 2 33218
33218 3 -1

题意：

链表的转置。给定N和K，每K个转一转，最后还要整个再转一转。

题解：

这道题挺熟悉的，和1074 Reversing Linked List (25分)差不多嘛。

当时依稀记得，用vector超方便，但是忘了怎么用

reverse(valid.begin()+i*k,valid.begin()+i*k+k);

然后就十分痛苦地用不了现成的reverse，只好把开两个vector当成数组用，然后非常艰难地写完了。。。。

AC代码：

#include<bits/stdc++.h>
using namespace std;
struct node{
    int d;
    int id;
    int nx;
}a[100005],b[100005];
vector<node>v;
int main(){
    int n,k;
    int root;
    cin>>root>>n>>k;
    for(int i=1;i<=n;i++){
        int x,y,z;
        cin>>x>>z>>y;
        a[x].d=z;
        a[x].nx=y;
        a[x].id=x;
    }
    node nroot=a[root];
    v.push_back(nroot);
    int num=1;
    while(nroot.nx!=-1){//先按照链表的顺序存好放到vector中 
        nroot=a[nroot.nx];
        v.push_back(nroot);
        num++;
    }
    int cs=num/k;//cs表示要转几次 
    for(int i=0;i<cs;i++){
        for(int j=0;j<k;j++){
            a[i*k+j]=v.at(i*k+k-j-1);//每k个转一转，本来可以 reverse(v.begin()+i*k,v()+i*k+k);唉。。。 
        }
    }
    int pp=cs*k-1;
    if(cs*k<num){//剩下的不到K个也要转 
        for(int i=v.size()-1;i>=cs*k;i--) a[++pp]=v.at(i);
    }
    for(int i=0;i<num;i++){//再整个链表转一下，本来可以reverse(v.begin(),v.end())的,唉。。。 
        b[i]=a[num-i-1];
    }
    for(int i=0;i<num;i++){//修改b数组里每个结构体的nx值 
        if(i!=num-1) b[i].nx=b[i+1].id;
        else b[i].nx=-1;
    }
    for(int i=0;i<num-1;i++){//输出 
        printf("%05d %d %05d
",b[i].id,b[i].d,b[i].nx);
    }
    printf("%05d %d -1",b[num-1].id,b[num-1].d);
    return 0;
}