Travelling（HDU3001+状压dp+三进制+最短路）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3001

题目：

题意：n个城市，m条边，每条边都有一个权值，问你经过所有的城市且每条边通过次数不超过两次的最短距离。

思路：状压dp+三进制，dp[i][j]表示在状态i下以j为目标城市的最短距离，转移方程为nw = i + fi[k];dp[nw][k] = min(dp[nw][k], dp[i][j] + mp[j][k])。

代码实现如下：

#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;;
typedef pair<int, int> pii;
typedef unsigned long long ull;

#define lson i<<1
#define rson i<<1|1
#define bug printf("*********
");
#define FIN freopen("D://code//in.txt", "r", stdin);
#define debug(x) cout<<"["<<x<<"]" <<endl;
#define IO ios::sync_with_stdio(false),cin.tie(0);

const double eps = 1e-8;
const int mod = 10007;
const int maxn = 60000 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f;

int n, m, u, v, w;
int mp[15][15], dp[maxn][15], fi[15], state[maxn][15];

void init() {
    fi[1] = 1;
    for(int i = 2; i <= 10; i++) {
        fi[i] = fi[i-1] * 3;
    }
    for(int i = 0; i < 60007; i++) {
        int t = i;
        for(int j = 1; j <= 10; j++) {
            state[i][j] = t % 3;
            t /= 3;
            if(t == 0) break;
        }
    }
}

int main() {
    //FIN;
    init();
    while(~scanf("%d%d", &n, &m)) {
        memset(mp, inf, sizeof(mp));
        memset(dp, inf, sizeof(dp));
        for(int i = 1; i <= n; i++) {
            mp[i][i] = 0, dp[fi[i]][i] = 0;
        }
        for(int i = 1; i <= m; i++) {
            scanf("%d%d%d", &u, &v, &w);
            mp[u][v] = mp[v][u] = min(mp[u][v], w);
        }
        int ans = inf;
        for(int i = 0; i < 3 * fi[n]; i++) {
            int flag = 1;
            for(int j = 1; j <= n; j++) {
                if(state[i][j] == 0) flag = 0;
                if(dp[i][j] >= inf) continue;
                for(int k = 1; k <= n; k++) {
                    if(j == k) continue;
                    if(mp[j][k] >= inf || state[i][k] >= 2) continue;
                    int nw = i + fi[k];
                    dp[nw][k] = min(dp[nw][k], dp[i][j] + mp[j][k]);
                }
            }
            if(flag) {
                for(int j = 1; j <= n; j++) {
                    ans = min(ans, dp[i][j]);
                }
            }
        }
        if(ans >= inf) printf("-1
");
        else printf("%d
", ans);
    }
    return 0;
}