uva 1347(双调欧几里得旅行商问题)

题意：一个人要去n个地方旅行，给出n个点的坐标，他会从最左边的点走到最右边，然后再回到起点，除了最左边和最右边两个点，每个点只经过一次。问最短路。

题解：f[i][j]表示从i到1再从1到j连通的最短路。

状态转移方程：

f[i][j] = f[i - 1][j] + dist(p[i - 1], p[i])

f[i][i - 1] = min(f[i][i - 1], f[i - 1][j] + dist(p[j], p[i]))

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int N = 105;
const double INF = 10000000000000;
struct Point {
int x, y;
}p
;
int n;
double f

;

double dist(const Point& a, const Point& b) {
return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

int main() {
while (scanf("%d", &n) == 1) {
for (int i = 0; i < n; i++)
scanf("%d%d", &p[i].x, &p[i].y);
for (int i = 0; i < n; i++)
for (int j = i; j < n; j++)
f[i][j] = f[j][i] = 0;
f[0][0] = 0;
for (int i = 1; i < n; i++) {
f[i][i - 1] = INF;
f[i][0] = dist(p[i], p[0]);
for (int j = 0; j < i - 1; j++) {
f[i][j] = f[i - 1][j] + dist(p[i], p[i - 1]);
f[i][i - 1] = min(f[i][i - 1], f[i - 1][j] + dist(p[i], p[j]));
}
}
printf("%.2lf\n", f[n - 1][n - 2] + dist(p[n - 1], p[n - 2]));
}
return 0;
}