FZU 1009 Jogging Trails 最短路+欧拉回路 解题报告

Problem Description

Gord is training for a marathon. Behind his house is a park with a large network of jogging trails connecting water stations. Gord wants to find the shortest jogging route that travels along every trail at least once.

Input consists of several test cases. The first line of input for each case contains two positive integers:n <= 15, the number of water stations, and m < 1000, the number of trails. For each trail, there is one subsequent line of input containing three positive integers: the first two, between 1 andn, indicating the water stations at the end points of the trail; the third indicates the length of the trail, in cubits. There may be more than one trail between any two stations; each different trail is given only once in the input; each trail can be travelled in either direction. It is possible to reach any trail from any other trail by visiting a sequence of water stations connected by trails. Gord's route may start at any water station, and must end at the same station. A single line containing 0 follows the last test case.

For each case, there should be one line of output giving the length of Gord's jogging route.

Sample Input

4 51 2 32 3 43 4 51 4 101 3 120

Sample Output

41 

题意：它让我们经过每条边至少一次，然后回到原点，求可以达到要求的最短的总路径。

所以，这个路径为欧拉回路，欧拉回路要求所有点的入度为偶数。

由于n<15，所以用dis[111111111]来记录状态，1为偶，0为奇。

代码（模仿大神代码写的，原文）：

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

const int maxn = 15+1;
const int maxm = 1<pos = pos;
        this->dist = dist;
    }
    //重载小于运算符
    bool operator<(const struct Node &ans)const
    {
        return dist > ans.dist;
    }
};

int degree[maxn],edge[maxn][maxn],dis[maxm];
bool vis[maxm];
int n, m, st, ed, sum;

void dijkstra()
{
    priority_queue Q;
    int up = 1 << n;
    for(int i = 0; i< up;i++)
    {
        dis[i] = -1;
        vis[i] = false;
    }
    dis[st] = sum;
    Q.push(Node(st,dis[st]));
    while(!Q.empty())
    {
        Node ans = Q.top();
        Q.pop();
        int u = ans.pos;
        if(vis[u]) continue;
        vis[u] = true;
        if(u == ed) break;
        for(int i = 0; i  c)
                edge[a][b] = edge[b][a] = c;
        }
        st = ed = 0;
        for(int i = 0; i < n; i++)
        {
            ed |= (1 << i);
            if(degree[i] % 2 == 0)
                st |= (1 << i);
        }
        dijkstra();
        printf("%d\n", dis[ed]);
    }
    return 0;
}