LightOJ - 1086 Jogging Trails（欧拉+状态压缩）

题目大意：有一个人要跑完所有的路，且要跑的路程最短，问如何跑

解题思路：跑完所有的路，且要跑的路程最短，跑欧拉路肯定是最短的。但是给出的图有可能不是欧拉回路，所以得自己再拼凑一下
无向图的欧拉回路就是所有点的度都是偶数了，所以找出所有度为奇数的点，状压求解连接这些点的最短路

#include 
#include 
#include 
using namespace std;
const int N = 20;
const int INF = 0x3f3f3f3f;
const int S = (1 << 15) + 10;

int dis[N][N];
int degree[N], dp[S];
int ans, n, m, cas = 1;

void floyd() {
    for (int k = 0; k < n; k++)
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                if (dis[i][k] != INF && dis[k][j] != INF && dis[i][j] > dis[i][k] + dis[k][j])
                    dis[i][j] = dis[i][k] + dis[k][j];
}

void init() {
    scanf("%d%d", &n, &m);
    memset(dis, 0x3f, sizeof(dis));
    for (int i = 0; i < n; i++) {
        dis[i][i] = 0;
        degree[i] = 0;
    }

    ans = 0;
    int u, v, d;
    for (int i = 0 ; i < m; i++)  {
        scanf("%d%d%d", &u, &v, &d);
        ans += d;
        u--; v--;
        dis[u][v] = dis[v][u] = min(d, dis[u][v]);
        degree[u]++;
        degree[v]++;
    }
    floyd();
}

int dfs(int state) {
    if (state == 0) return 0;
    if (~dp[state]) return dp[state];

    int s;
    for (int i = 0; i < n; i++) {
        if (state & (1 << i)) {
            s = i;
            break;
        }
    }

    dp[state] = INF;
    for (int i = s + 1; i < n; i++)
        if (state & (1 << i)) 
            dp[state] = min(dp[state], dfs(state ^ (1 << s) ^ (1 << i)) + dis[s][i]);

    return dp[state];
}

void solve() {
    int state = 0;
    for (int i = 0; i < n; i++)
        if (degree[i] % 2)  {
            state |= (1 << i);
        }
    memset(dp, -1, sizeof(dp));
    ans += dfs(state);
    printf("Case %d: %d\n", cas++, ans);
}

int main() {
    int test;
    scanf("%d", &test);
    while (test--) {
        init();
        solve();
    }
    return 0;
}