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

int dis

;
int degree
, 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;
}