1150 Travelling Salesman Problem (25分)
The "travelling salesman problem" asks the following question: "Given a list of cities and the distances between each pair of cities, what is the shortest possible route that visits each city and returns to the origin city?" It is an NP-hard problem in combinatorial optimization, important in operations research and theoretical computer science. (Quoted from "https://en.wikipedia.org/wiki/Travelling_salesman_problem".)
In this problem, you are supposed to find, from a given list of cycles, the one that is the closest to the solution of a travelling salesman problem.
Input Specification:
Each input file contains one test case. For each case, the first line contains 2 positive integers N (2<N≤200), the number of cities, and M, the number of edges in an undirected graph. Then M lines follow, each describes an edge in the format City1 City2 Dist, where the cities are numbered from 1 to N and the distance Dist is positive and is no more than 100. The next line gives a positive integer K which is the number of paths, followed by K lines of paths, each in the format:
n C1 C2 ... Cn
where n is the number of cities in the list, and Ci's are the cities on a path.
Output Specification:
For each path, print in a line Path X: TotalDist (Description) where X is the index (starting from 1) of that path, TotalDist its total distance (if this distance does not exist, output NA instead), and Description is one of the following:
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.
Finally print in a line Shortest Dist(X) = TotalDist where X is the index of the cycle that is the closest to the solution of a travelling salesman problem, and TotalDist is its total distance. It is guaranteed that such a solution is unique.
Sample Input:
6 10
6 2 1
3 4 1
1 5 1
2 5 1
3 1 8
4 1 6
1 6 1
6 3 1
1 2 1
4 5 1
7
7 5 1 4 3 6 2 5
7 6 1 3 4 5 2 6
6 5 1 4 3 6 2
9 6 2 1 6 3 4 5 2 6
4 1 2 5 1
7 6 1 2 5 4 3 1
7 6 3 2 5 4 1 6
Sample Output:
Path 1: 11 (TS simple cycle)
Path 2: 13 (TS simple cycle)
Path 3: 10 (Not a TS cycle)
Path 4: 8 (TS cycle)
Path 5: 3 (Not a TS cycle)
Path 6: 13 (Not a TS cycle)
Path 7: NA (Not a TS cycle)
Shortest Dist(4) = 8
思路
当满足将图中所有点也就是n个点都遍历一次， 且开始的点和终点也遍历一次且每个点都联通 就满足了 TS cycle, 当遍历的总数恰好是n + 1 那么满足TS simple cycle 否则是 Not a TS cycle
遍历所有的点可以用 set来判断,当set中存放的点是n时 说明所有的点都遍历,
代码
#include<bits/stdc++.h>
using namespace std;
const int maxsize = 305;
int edge[maxsize][maxsize];
int main()
{
int n, m;
scanf("%d%d", &n, &m);
int from, to, cost;
for(int i = 0; i < m; i++) {
scanf("%d%d%d", &from, &to, &cost);
edge[from][to] = edge[to][from] = cost;
}
int k, cnt = 1, sum, start, mins = 99999999, pos = -1;
scanf("%d", &k);
while(k--) {
scanf("%d", &sum);
set<int> s;
vector<int> v;
int t, count = 0;
bool srepeat = false, isexist = true;
v.resize(sum);
for(int i = 0; i < sum; i++) { // 此处可以和原来的代码进行对比
scanf("%d", &v[i]);
s.insert(v[i]); // 在输入点时就可以将值加入set
}
for(int i = 0; i < sum - 1; i++) { // 此处在单独判断点的连通性
if(edge[v[i]][v[i + 1]] == 0) isexist = false;
count += edge[v[i]][v[i + 1]];
}
printf("Path %d: ", cnt);
if(isexist == false) printf("NA (Not a TS cycle)
");
else{
printf("%d ", count);
if(s.size() == n && v[0] == v[sum - 1]) {
if(v.size() == n + 1) {
if(count < mins) {
mins = count;
pos = cnt;
}
printf("(TS simple cycle)
");
} else{
if(count < mins) {
mins = count;
pos = cnt;
}
printf("(TS cycle)
");
}
} else {
printf("(Not a TS cycle)
");
}
}
cnt++;
}
printf("Shortest Dist(%d) = %d
", pos, mins);
return 0;
}