UVa 10911 Forming Quiz Teams(状态压缩DP)
2012-11-22 13:35
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题意:
有2*n个点,使其组成n对,求n对点集的最小距离之和。
思路:
由于2*n最大才20,完全可以由数字的位来表示,每一个数字表示一个状态,然后才去记忆化搜索的方式得出结果。
(看了别人的题解才想起来怎么去做,关于状态压缩的动归,还是没能很好的形成思维。多观察,多总结规律然后归纳之,减少冗余,使收益最大化)
有2*n个点,使其组成n对,求n对点集的最小距离之和。
思路:
由于2*n最大才20,完全可以由数字的位来表示,每一个数字表示一个状态,然后才去记忆化搜索的方式得出结果。
(看了别人的题解才想起来怎么去做,关于状态压缩的动归,还是没能很好的形成思维。多观察,多总结规律然后归纳之,减少冗余,使收益最大化)
#include <cstdio> #include <cstdlib> #include <cstring> #include <climits> #include <algorithm> #include <cmath> using namespace std; const int MAXD = 65600; const int MAXN = 25; double map[MAXN][MAXN]; int x[MAXN], y[MAXN], n; double dp[MAXD]; double solve(int s) { if (dp[s] != -1) return dp[s]; dp[s] = INT_MAX; for (int i = 0; i < n; ++i) if (s & (1 << i)) for (int j = 0; j < n; ++j) if (i != j && (s & (1 << j))) dp[s] = min(dp[s], solve(s ^ (1 << i) ^ (1 << j)) + map[i][j]); return dp[s]; } int main() { int cases = 0; while (scanf("%d", &n) && n) { char b[100]; n <<= 1; for (int i = 0; i < n; ++i) scanf("%s %d %d", b, &x[i], &y[i]); for (int i = 0; i < n; ++i) for (int j = 0; j < i; ++j) map[i][j] = map[j][i] = sqrt(1.0 * ((x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]))); int end = (1 << n) - 1; for (int i = 1; i <= end; ++i) dp[i] = -1; dp[0] = 0; solve(end); printf("Case %d: %.2f\n", ++cases, dp[end]); } return 0; }
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