1074 Reversing Linked List (25分)
Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K=3, then you must output 3→2→1→6→5→4; if K=4, you must output 4→3→2→1→5→6.
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 (≤105) which is the total number of nodes, and a positive K (≤N) which is the length of the sublist to be reversed. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.
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 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218
Sample Output:
00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1
第五测试点:答案错误原因是 存在一些不在头结点引出的链表上的结点。
我的代码：
收获：cout输出string可能超时
可以用printf 输出 格式：str.c_str()
#include<iostream>
#include<map>
#include<algorithm>
#include<vector>
#include<cstring>
using namespace std;
struct node{
string ad;
int d;
string next;
};
map<string,node>::iterator ip;
int main()
{
map<string,node> s;
vector<node> a,b;
b.clear();
a.clear();
s.clear();
int i,j,t,n,len;
string add,k;
node temp;
cin>>add>>n>>t;
for(i=0;i<n;i++)
{
cin>>temp.ad>>temp.d>>temp.next;
s.insert(make_pair(temp.ad,temp));
}
for(i=0;i<n;i++)
{
if(s.count(add))
{
ip=s.find(add);
a.push_back(ip->second);
add=ip->second.next;
// cout<<1<<endl;
}
}
n=a.size();
for(i=t;i<=n;)
{
for(j=1;j<=t;j++)
{
// cout<<a[i-j].ad<<" "<<a[i-j].d<<" "<<a[i-j].next<<endl;
b.push_back(a[i-j]);
}
i+=t;
}
if(i!=n)
{
i-=t;
for(j=i;j<n;j++)
{
b.push_back(a[j]);
}
}
len=b.size()-1;
for(i=0;i<len;i++)
{
printf("%05s %d %05s\n",b[i].ad.c_str(),b[i].d,b[i+1].ad.c_str());
// cout<<b[i].ad<<" "<<b[i].d<<" "<<b[i+1].ad<<endl;
}
// cout<<b[len].ad<<" "<<b[b.size()-1].d<<" -1"<<endl;
printf("%05s %d -1\n",b[len].ad.c_str(),b[len].d);
return 0;
}别人的代码：
#include <iostream>
#include<fstream>
using namespace std;
int main() {
int h, k, n, sum = 0;
#ifdef _DEBUG
ifstream cin("data.txt");
#endif
cin >> h >> n >> k;
int temp, data[100005], next[100005], list[100005], result[100005];
for (int i = 0; i < n; i++) {
cin >> temp;
cin >> data[temp] >> next[temp];
}
while (h != -1) {
list[sum++] = h;
h = next[h];
}
for (int i = 0; i < sum; i++) result[i] = list[i];
for (int i = 0; i < (sum - sum % k); i++)
result[i] = list[i / k * k + k - 1 - i % k];
for (int i = 0; i < sum - 1; i++)
printf("%05d %d %05d\n", result[i], data[result[i]], result[i + 1]);
printf("%05d %d -1", result[sum - 1], data[result[sum - 1]]);
#ifdef _DEBUG
cin.close();
#endif
return 0;
}
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
const int maxn = 1e5+100;
int a[maxn],c[maxn];
int b[maxn][2];
int main()
{
int digit,n,m;
scanf("%d%d%d",&digit,&n,&m);
for(int i = 1; i <= n; i++)
{
int x,y,z;
scanf("%d%d%d",&x, &y, &z);
b[x][0] = y;
b[x][1] = z;
}
int gg = 0;
a[++gg] = digit;
for(++gg;a[gg-1] != -1;++gg)
a[gg] = b[a[gg-1]][1];
gg--;gg--;
// cout<<gg<<endl;
// for(int i =1 ; i<=gg;i++)
// cout<<a[i]<<endl;
int now = 0,i;
for(i = m;i <= gg; i += m )
{
for(int j = i; j > i-m; j--){
c[++now] = a[j];
}
}
if(i>gg){
for(i = i- m + 1; i <= gg;i++)
c[++now] = a[i];
}
for(int i = 1 ; i < gg;i++)
printf("%05d %d %05d\n",c[i],b[c[i]][0],c[i+1]);
printf("%05d %d %d\n",c[gg],b[c[gg]][0],-1);
return 0;
} #include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn = 1e5 + 10;
struct Node
{
int add, data, next;
}node[maxn];
void init()
{
for (int i = 0; i < maxn; i++)node[i].add = i;
}
int head, n, k;
vector<Node>list;
int main()
{
scanf("%d%d%d", &head, &n, &k);
init();
for (int i = 0; i < n; i++)
{
int address;
scanf("%d", &address);
scanf("%d%d", &node[address].data, &node[address].next);
}
int p = head;
while (p != -1)
{
list.push_back(node[p]);
p = node[p].next;
}
int group = list.size() / k;
for (int i = 0; i < group; i++)
{
reverse(list.begin() + i*k, list.begin() + i*k + k);
}
for (int i = 0; i < list.size(); i++)
{
printf("%05d %d ", list[i].add, list[i].data);
if (i != list.size() - 1)printf("%05d", list[i + 1].add);
else printf("-1");
printf("\n");
}
return 0;
}