正题

luogu 6085

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题目大意

给你一个无向图，其中有一些边是必须走的，问你从1开始走，经过所有必须走的边，然后回到1的最短路径

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解题思路

n很小，可以先用Floyd跑出两个点之间的最短路

然后状压DP，每个点存三种状态：
0.不在连通图内
1.在连通图内且连边为奇数
2.在连通图内且连边为偶数

遍历所有状态，每个状态找不在连通图内的点，然后和连通图内一点连边，然后记录奇偶性（必要的边不记录）

枚举完所有状态后，把必要的边加进里面，然后对于连边为奇数的，找到一种最小代价连接方式，这样可以构造一条回路

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代码

#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define N 1594323
using namespace std;
int n, m, k, x, y, z, p, MX, nec, sum, ans, tot, h[100], v[100], g[10000], deg[100], dis[100][100];
int f[N + 10];
queue<int>d;
struct rec
{int to, l, next;
}e[200];
void add(int x, int y, int z)
{e[++tot].to = y;e[tot].l = z;e[tot].next = h[x];h[x] = tot;
}
int main()
{memset(dis, 127/3, sizeof(dis));memset(g, 127/3, sizeof(g));memset(f, 127/3, sizeof(f));MX = g[0];scanf("%d%d", &n, &m);for (int i = 1; i <= m; ++i){scanf("%d%d%d", &x, &y, &z);x--;y--;add(x, y, z);add(y, x, z);dis[x][y] = dis[y][x] = min(dis[x][y], z);nec ^= (1<<x) ^ (1<<y);deg[x]++;deg[y]++;sum += z;}scanf("%d", &m);for (int i = 1; i <= m; ++i){scanf("%d%d%d", &x, &y, &z);x--;y--;dis[x][y] = dis[y][x] = min(dis[x][y], z);}v[0] = 1;dis[0][0] = 0;for (int i = 1; i <= n; ++i)v[i] = v[i - 1] * 3, dis[i][i] = 0;for (int k = 0; k < n; ++k)for (int i = 0; i < n; ++i)for (int j = 0; j < n; ++j)dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);//Floydg[0] = 0;for (int now = 0; now < (1<<n); ++now)for (int i = 0; i < n; ++i)if (!(now & (1<<i)))for (int j = i + 1; j < n; ++j)if (!(now & (1<<j))){k = now ^ (1<<i) ^ (1<<j);g[k] = min(g[k], g[now] + dis[i][j]);//预处理奇数点的连边}d.push(2);f[2] = 0;while(!d.empty()){int now = d.front();d.pop();for (int i = 0; i < n; ++i){if (now / v[i] % 3) continue;k = now + v[i] * 2;//必须的连边不算奇偶性for (int j = h[i]; j; j = e[j].next){if (!(now / v[e[j].to] % 3)) continue;if (f[k] >= MX) d.push(k);f[k] = min(f[k], f[now]);}for (int j = 0; j < n; ++j)//不必须的连边{if (!(now / v[j] % 3)) continue;k = now + v[i];if (now / v[j] % 3 == 2) k -= v[j];//计算奇偶性else k += v[j];if (f[k] >= MX) d.push(k);f[k] = min(f[k], f[now] + dis[j][i]);}}}ans = MX;for (int now = 0; now <= v[n]; ++now){p = 0;k = 0;for (int i = 0; i < n; ++i){if (!(now / v[i] % 3) && deg[i])//有必须的点不在连通图内{p = 1;break;}if (now / v[i] % 3) k ^= (1<<i) * (now / v[i] % 3 == 1? 1: 0);//计算奇偶性}if (p) continue;k ^= nec;ans = min(ans, f[now] + g[k]);}printf("%d", sum + ans);return 0;
}