uva 1347 - Tour(双调欧几里得旅行商问题）

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题目链接：uva 1347 - Tour

题目大意：给出n个点，确定一条连接各点的最短闭合旅程的问题。

解题思路：dp[i][j]表示说从i联通到1，再从1联通到j的距离。

dp[i][j] = dp[i-1][j] + dis(i,i-1);

dp[i][i-1] = min (dp[i][i-1], dp[i-1][j] + dis(i, j));

双调欧几里得旅行商问题

#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std;const int N = 105;const double INF = 0x3f3f3f3f3f3f3f3f;int n;double x[N], y[N], dp[N][N];inline double dis (int a, int b) {return sqrt((x[a]-x[b])*(x[a]-x[b]) + (y[a]-y[b])*(y[a]-y[b]));}void init () {for (int i = 1; i <= n; i++)scanf("%lf%lf", &x[i], &y[i]);memset(dp, 0, sizeof(dp));dp[2][1] = dis(1, 2);}double solve () {for (int i = 3; i <= n; i++) {dp[i][i-1] = INF;for (int j = 1; j < i-1; j++) {dp[i][i-1] = min(dp[i][i-1], dp[i-1][j] + dis(i, j));dp[i][j] = dp[i-1][j] + dis(i, i-1);}}double ans = INF;for (int i = 1; i < n; i++)ans = min(ans, dp[n][i] + dis(n, i));return ans;}int main () {while (scanf("%d", &n) == 1) {init ();printf("%.2lf\n", solve ());}return 0;}

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