【PAT甲级题解记录】1150 Travelling Salesman Problem (25 分)

【PAT甲级题解记录】1150 Travelling Salesman Problem (25 分)

前言

Problem：1150 Travelling Salesman Problem (25 分)

Tags：模拟 图的遍历 旅行商问题

Difficulty：剧情模式 想流点汗 想流点血 死而无憾

Address：1150 Travelling Salesman Problem (25 分)

问题描述

给定一个图和一些路径，求这些路径是否为旅行商环路、简单旅行商环路或者非旅行商环路。

• TS simple cycle if it is a simple cycle that visits every city;
• TS cycle if it is a cycle that visits every city, but not a simple cycle;
• Not a TS cycle if it is NOT a cycle that visits every city.

解题思路

只要会存储图，模拟一下就可以了。

唯一的难点是三种路径类型的判断上并没有那么轻松，甚至读题也有点麻烦，还是有点烦的，不过样例能过基本上就没问题了。

建议在遍历路径判断前像这样打个草稿，这种题思路一定要严谨清晰。

// Not a TS cycle if it is NOT a cycle that visits every city.
if(不连通 or 首尾不通 or 不包含全部) // 其中不连通输出NA// TS cycle if it is a cycle that visits every city, but not a simple cycle;
else if(有重复): // TS simple cycle if it is a simple cycle that visits every city;
else: 

参考代码

#include<iostream>
#include<cstdio>
#include<vector>
#include<set>using namespace std;
int N;  // the number of cities
int M;  // the number of edges in an undirected graph
vector<vector<int>> edges;  // 邻接矩阵
void init() {cin >> N >> M;// 初始化 edges (-1)edges.resize(N + 1);for (int i = 1; i <= N; ++i) {edges[i].resize(N + 1, -1);}// 输入 edgesfor (int i = 0; i < M; ++i) {int c1, c2, d;cin >> c1 >> c2 >> d;edges[c1][c2] = d;edges[c2][c1] = d;}}void solve() {int K;cin >> K;int min_dist = 0x3f3f3f3f;int min_index = -1;for (int k = 1; k <= K; ++k) {int n;cin >> n;vector<int> path(n);for (int i = 0; i < n; i++) {cin >> path[i];}// checkint dist = 0;bool is_conn = true; // 判断路径通不通bool is_all = true; // 判断是否 visit every citybool is_simple = true; // 判断重复（是否简单环）set<int> visited; // 访问计数，用来判断重复visited.insert(path[0]);for (int i = 1; i < n; ++i) {if (edges[path[i - 1]][path[i]] == -1) {is_conn = false;}if (i != n - 1 && visited.count(path[i])) {  // 注意末尾不判断，末尾是用来构成环的is_simple = false;}dist += edges[path[i - 1]][path[i]];visited.insert(path[i]);}if (visited.size() != N) {is_all = false;}if (!is_conn) {printf("Path %d: NA (Not a TS cycle)\n", k);} else if (!is_all || path[0] != path[n - 1]) {printf("Path %d: %d (Not a TS cycle)\n", k, dist);} else if (!is_simple) {printf("Path %d: %d (TS cycle)\n", k, dist);if (dist < min_dist) {min_dist = dist;min_index = k;}} else {printf("Path %d: %d (TS simple cycle)\n", k, dist);if (dist < min_dist) {min_dist = dist;min_index = k;}}}printf("Shortest Dist(%d) = %d\n", min_index, min_dist);
}void solution_1150() {init();solve();
}
int main() {solution_1150();return 0;
}