uva 10296 - Jogging Trails （中国邮路问题 状压dp）

Problem B: Jogging Trails

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 and n, 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 5
1 2 3
2 3 4
3 4 5
1 4 10
1 3 12
0

Output for Sample Input

41

从起点出发，经过所有边至少一次后回到起点，求最短路径。网上很多分析，我就不写了。

#include<cstdio>
#include<map>
#include<queue>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<list>
#include<set>
#include<cmath>
using namespace std;
const int maxn = 100 + 5;
const int INF = 1e9;
const double eps = 1e-6;
typedef unsigned long long ULL;
typedef long long LL;
typedef pair<int, int> P;
#define fi first
#define se second

int n, m;
int dis[maxn][maxn];

void floyd(){
    for(int k = 1;k <= n;k++){
        for(int i = 1;i <= n;i++){
            for(int j = 1;j <= n;j++)
                dis[i][j] = min(dis[i][j], dis[i][k]+dis[k][j]);
        }
    }
}

int degree[maxn], dp[1<<20];

int solve(int now){
    if(now == 0)
        return 0;
    if(dp[now] != -1)
        return dp[now];
    int ret = INF;
    for(int i = 1;i <= n;i++){
        if(now&(1<<(i-1))){
            for(int j = i+1;j <= n;j++){
                if(now&(1<<(j-1))){
                    ret = min(ret, dis[i][j]+solve(now-(1<<(i-1))-(1<<(j-1))));
                }
            }
        }
    }
    return dp[now] = ret;
}

int main(){
    while(cin >> n){
        if(n == 0)
            break;
        cin >> m;
        memset(degree, 0, sizeof degree);
        for(int i = 1;i <= n;i++){
            for(int j = 1;j <= n;j++)
                dis[i][j] = INF;
            dis[i][i] = 0;
        }
        int ans = 0;
        while(m--){
            int x, y, c;
            cin >> x >> y >> c;
            dis[x][y] = dis[y][x] = min(dis[x][y], c);
            degree[x]++;
            degree[y]++;
            ans += c;
        }
        floyd();
        int ori = 0;
        for(int i = 1;i <= n;i++){
            ori += ((degree[i]&1)<<(i-1));
        }
        memset(dp, -1, sizeof dp);
        ans += solve(ori);
        cout << ans << endl;
    }
    return 0;
}