Hdu 2224 The shortest path（双调TSP）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2224 

The shortest path

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 654    Accepted Submission(s): 335


[align=left]Problem Description[/align]
There are n points on the plane, Pi(xi, yi)(1 <= i <= n), and xi < xj (i<j). You begin at P1 and visit all points then back to P1. But there is a constraint: 

Before you reach the rightmost point Pn, you can only visit the points those have the bigger x-coordinate value. For example, you are at Pi now, then you can only visit Pj(j > i). When you reach Pn, the rule is changed, from now on you can only visit the points
those have the smaller x-coordinate value than the point you are in now, for example, you are at Pi now, then you can only visit Pj(j < i). And in the end you back to P1 and the tour is over.

You should visit all points in this tour and you can visit every point only once.
 

[align=left]Input[/align]
The input consists of multiple test cases. Each case begins with a line containing a positive integer n(2 <= n <= 200), means the number of points. Then following n lines each containing two positive integers Pi(xi, yi),
indicating the coordinate of the i-th point in the plane.
 

[align=left]Output[/align]
For each test case, output one line containing the shortest path to visit all the points with the rule mentioned above.The answer should accurate up to 2 decimal places.
 

[align=left]Sample Input[/align]

3
1 1
2 3
3 1

 

[align=left]Sample Output[/align]

6.47

Hint: The way 1 - 3 - 2 - 1 makes the shortest path.

 

题意：给n个二维点，你要先沿x坐标严格递增的方向从1点走到n点，然后在沿x坐标严格递减的方向从n点走到1点，求走过的路程和的最小值。

分析：可以将上述的一条路看成从1点到n点的没有交叉点的两条路A和B。 两条路分别从 点1 向x递增的方向 A走到 i，B走到 j 。

就可以定义 dp[i][j] 为A走到 i ，B走到 j 的最短距离。题目就变成了 求dp

的值。

转移方程：

dp[1][2] = D[i][j]; //i ,j 两点的真实距离。

i < j-1 时； dp[ i ][ j ] = dp[ i ][ j-1] + D[ j-1 ][ j ];

i == j-1 时；dp[ i ][ j ] = min(dp[ k ][ j-1 ] + D[ k ][ j ]);

dp[ n ][ n ] = dp[ n-1 ][ n ]+ D[ n-1 ][ n ].

代码：

  #include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;
#define MAX 205
#define inf 9999999
double dp[MAX][MAX];
double D[MAX][MAX];
struct node
{
double x,y;
}p[MAX];
int cmp(const void *a,const void *b)
{
struct node *p = (struct node *)a;
struct node *q = (struct node *)b;
return (p->x - q->x >0)?1:-1;
}
double dis(node a,node b)
{
return sqrt(pow(a.x-b.x,2)+pow(a.y-b.y,2));
}
void init(int n)
{
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
{
D[i][j] = inf;
}
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
init(n);
for (int i = 1;i<=n;i++)
{
scanf("%lf %lf",&p[i].x,&p[i].y);
for(int j=1;j<i;j++)
D[i][j]=D[j][i] = dis(p[i],p[j]);//记录直接距离。
}
qsort(p+1,n,sizeof(p[1]),cmp);
dp[1][2] = D[1][2];
for (int i = 3; i <= n; i++)
{
for (int j = 1; j <= i-2; j++)//when i-1 in the higher dir
{
dp[j][i] = dp[j][i-1] + D[i-1][i];
}
double Min = inf;
for (int j = 1; j <= i-2; j++)//when i-1 in the lower dir
{
Min = min(dp[j][i-1]+D[j][i],Min);
}dp[i-1][i] = Min;
}
dp

= dp[n-1]
+D
[n-1];
printf("%.2lf\n",dp

);
}
return 0;
}