算法笔记 上机训练实战指南 第7章 提高篇(1)--数据结构专题(1) 学习笔记

Mice and Rice is the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. The goal of each mouse is to eat as much rice as possible in order to become a FatMouse.

First the playing order is randomly decided for N?P?? programmers. Then every N?G?? programmers are grouped in a match. The fattest mouse in a group wins and enters the next turn. All the losers in this turn are ranked the same. Every N?G?? winners are then grouped in the next match until a final winner is determined.

For the sake of simplicity, assume that the weight of each mouse is fixed once the programmer submits his/her code. Given the weights of all the mice and the initial playing order, you are supposed to output the ranks for the programmers.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers: N?P?? and N?G?? (≤), the number of programmers and the maximum number of mice in a group, respectively. If there are less than N?G?? mice at the end of the player‘s list, then all the mice left will be put into the last group. The second line contains N?P?? distinct non-negative numbers W?i?? (,) where each W?i?? is the weight of the i-th mouse respectively. The third line gives the initial playing order which is a permutation of 0 (assume that the programmers are numbered from 0 to N?P??−1). All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the final ranks in a line. The i-th number is the rank of the i-th programmer, and all the numbers must be separated by a space, with no extra space at the end of the line.

Sample Input:

11 3
25 18 0 46 37 3 19 22 57 56 10
6 0 8 7 10 5 9 1 4 2 3

Sample Output:

5 5 5 2 5 5 5 3 1 3 5

Given a singly linked list L with integer keys, you are supposed to remove the nodes with duplicated absolute values of the keys. That is, for each value K, only the first node of which the value or absolute value of its key equals K will be kept. At the mean time, all the removed nodes must be kept in a separate list. For example, given L being 21→-15→-15→-7→15, you must output 21→-15→-7, and the removed list -15→15.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, and a positive N (≤) which is the total number of nodes. 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 Key Next

where Address is the position of the node, Key is an integer of which absolute value is no more than 1, and Next is the position of the next node.

Output Specification:

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

Sample Input:

00100 5
99999 -7 87654
23854 -15 00000
87654 15 -1
00000 -15 99999
00100 21 23854

Sample Output:

00100 21 23854
23854 -15 99999
99999 -7 -1
00000 -15 87654
87654 15 -1

#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 100010;
const int TABLE = 1000010; 
struct Node{
    int address,data,next;
    int order;
}node[maxn];
bool isExist[TABLE] = {false};
bool cmp(Node a,Node b){
    return a.order < b.order;
}
int main(){
    memset(isExist,false,sizeof(isExist));
    for(int i=0;i<maxn;i++){
        node[i].order = 2 * maxn;
    }
    int begin,n;
    scanf("%d%d",&begin,&n);
    int address;
    for(int i=0;i<n;i++){
        scanf("%d",&address);
        scanf("%d %d",&node[address].data,&node[address].next);
        node[address].address = address;
    }    
    int p = begin,countValid=0,countRemoved=0;
    while(p != -1){
        if(!isExist[abs(node[p].data)]){
            isExist[abs(node[p].data)] = true;
            node[p].order = countValid++;
        }else{
            node[p].order =  maxn +countRemoved++;
        }
        p = node[p].next;
    }
    int count = countRemoved + countValid;
    sort(node,node+maxn,cmp);
    for(int i=0;i<count;i++){
        if(i != countValid-1 && i != count -1){
            printf("%05d %d %05d\n",node[i].address,node[i].data,node[i+1].address);
        }else{
            printf("%05d %d -1\n",node[i].address,node[i].data);
        }
    }
    return 0;
}