UVa 10911 Forming Quiz Teams [DP]

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Description

最优配对问题
2n个点，配成n组，使得距离和最短、

Algorithm

动态规划

O(n^2*2^n）方法

dp[s] 表示已经配对了的集合为s的情况的最小距离和
每次枚举s,i,j

Code

O(n^2*2^n）方法

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <climits>using  namespace std;int n;const int maxn = 20;struct Node {    int x, y;};Node nodes[maxn];double d[maxn][maxn];double dp[1<<maxn];double dis(Node node1, Node node2) {  return sqrt((node1.x - node2.x)*(node1.x - node2.x) + (node1.y - node2.y)*(node1.y - node2.y));}void solve() {  memset(d, 0, sizeof(d));  memset(dp, 0, sizeof(dp));  n *= 2;  int e = 1 << n; //全部集合  for (int i = 0; i < n; i++) {    string string1;    cin >> string1 >> nodes[i].x >> nodes[i].y;  }  for (int i = 0; i < n; i++)    for (int j = i + 1; j < n; j++) {      d[i][j] = dis(nodes[i], nodes[j]);    }  for (int i = 1; i < e; i++) {    dp[i] = INT_MAX;  }  for (int s = 0; s < e; s++) { //全部集合    for (int i = 0; i < n; i++) {      for (int j = i + 1; j < n; j++) {        int sij = (1 << i) | (1 << j); //i,j并集        if (!(s&sij)) { //i, j不在集合中          dp[s|sij] = min(dp[s|sij], dp[s] + d[i][j]);        }      }    }  }  printf("%.2lf\n", dp[e-1]);}int main() {  //freopen("input.txt", "r", stdin);  for (int i = 1;;i++) {    scanf("%d\n", &n);    if (n == 0) break;    printf("Case %d: ", i);    solve();  }}

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