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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 (≤10​5​​) 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 400000 4 9999900100 1 1230968237 6 -133218 3 0000099999 5 6823712309 2 33218

Sample Output:

00000 4 3321833218 3 1230912309 2 0010000100 1 9999999999 5 6823768237 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
   
    #include
    
   

别人的代码：

#include 
   
    #include
    
     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
   
    #include
    
     #include
     
      #include
      
       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; 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
   
    #include
    
     #include
     
      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
      
       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;}
      
     
    
   

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