題目連結
For every $dp[i] = k$, we define $i$ as the collection of people that’re not paired yet, and $k$ is the minimum cost to pair those people up, if possible.
Starting with (1 << n) - 1, we can pair 2 of them at a time, and recurse on (state ^ (1 << i) ^ (1 << j)).
Notice the base case is $dp[0] = 0$.
AC Code
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> ii;
const double oo = 1e9;
double dist[22][22];
void dfs(int s, double dp[], int n)
{
if (dp[s] != oo)
return;
// 1 means not paired uet
for (int i = 0; i < n; i++) {
if ((s >> i) & 1) {
for (int j = i + 1; j < n; j++) {
if (i == j) // pair with self, cont.
continue;
if ((s >> j) & 1) {
int nxt = s ^ (1 << i) ^ (1 << j);
dfs(nxt, dp, n);
dp[s] = min(dp[s], dp[nxt] + dist[i][j]);
}
}
}
}
}
double solve(int n)
{
n *= 2;
ii pos[n];
for (int i = 0; i < n; i++) {
char inp[111];
scanf("%s %d %d", inp, &pos[i].first, &pos[i].second);
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
int dx = pos[i].first - pos[j].first;
int dy = pos[i].second - pos[j].second;
dist[i][j] = sqrt(dx * dx + dy * dy);
}
}
// for (int i = 0; i < n; i++)
// for (int j = 0; j < n; j++) {
// printf("%7.2f%c", dist[i][j], j == n - 1 ? '\n' : ' ');
// }
double dp[1 << n];
fill(dp, dp + (1 << n), oo);
dp[0] = 0; // base case
dfs((1 << n) - 1, dp, n);
return dp[(1 << n) - 1];
}
int main()
{
int n;
for (int i = 1; scanf("%d", &n) == 1 && n; i++) {
printf("Case %d: %.2f\n", i, solve(n));
}
return 0;
}
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