PAT甲级 1097

题目
解析
代码

题目

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 (≤10^5 ) which is the total number of nodes. 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 Key Next
where Address is the position of the node, Key is an integer of which absolute value is no more than 10^4 , 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.

解析

输入包含address‘、key、next的数据，将该数据清理干净。key不能重复，也不能和key的绝对值重复。

代码

#include<bits/stdc++.h>#define INF 1<<29using namespace std;struct node {string address;int key;string next;
};
bool flag[10010] = {false};void pat1097() {string begin;int n;map<string, node> data;vector<node> output, remove;cin >> begin >> n;for (int i = 0; i < n; ++i) {string address, next;int key;cin >> address >> key >> next;data[address] = {address, key, next};}while (begin != "-1") {if (!flag[abs(data[begin].key)]) {output.push_back(data[begin]);flag[abs(data[begin].key)] = true;} else {remove.push_back(data[begin]);}begin = data[begin].next;}for (int i = 0; i < output.size(); ++i) {printf("%s %d ", output[i].address.c_str(), output[i].key);if (i != output.size() - 1)printf("%s\n", output[i + 1].address.c_str());elseprintf("-1\n");}for (int i = 0; i < remove.size(); ++i) {printf("%s %d ", remove[i].address.c_str(), remove[i].key);if (i != remove.size() - 1)printf("%s\n", remove[i + 1].address.c_str());elseprintf("-1\n");}
}int main() {pat1097();return 0;
}

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PAT甲级 1097

PAT甲级 1097

题目

解析

代码

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