POJ 2677 旅行商问题 双调dp或者费用流
Tour
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 3408 | Accepted: 1513 |
Description
John Doe, a skilled pilot, enjoys traveling. While on vacation, he rents a small plane and starts visiting beautiful places. To save money, John must determine the shortest closed tour that connects his destinations. Each destination is represented by a point in the plane pi = < xi,yi >. John uses the following strategy: he starts from the leftmost point, then he goes strictly left to right to the rightmost point, and then he goes strictly right back to the starting point. It is known that the points have distinct x-coordinates.
Write a program that, given a set of n points in the plane, computes the shortest closed tour that connects the points according to John's strategy.
Write a program that, given a set of n points in the plane, computes the shortest closed tour that connects the points according to John's strategy.
Input
The program input is from a text file. Each data set in the file stands for a particular set of points. For each set of points the data set contains the number of points, and the point coordinates in ascending order of the x coordinate. White spaces can occur freely in input. The input data are correct.
Output
For each set of data, your program should print the result to the standard output from the beginning of a line. The tour length, a floating-point number with two fractional digits, represents the result. An input/output sample is in the table below. Here there are two data sets. The first one contains 3 points specified by their x and y coordinates. The second point, for example, has the x coordinate 2, and the y coordinate 3. The result for each data set is the tour length, (6.47 for the first data set in the given example).
Sample Input
3 1 1 2 3 3 1 4 1 1 2 3 3 1 4 2
Sample Output
一旅行商从左向右走到最右边，然后再返回原来出发点的最短路径。
两种做法，第一种dp，dp[i][j]表示以i，j结尾的两条不相交的路径假设i一定大于j，i有两种选择，与i-1相连，不与i-1相连，然后dp
代码：
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define ll int
#define N 1005
#define inf 100000000
struct node{
double x, y;
bool operator<(const node&a)const{
if(a.x==x)return a.y>y;
return a.x>x;
}
}p[N];
double Dis(node a, node b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}
int n;
double dis[N][N],dp[N][N];
int main(){
ll i, j, u, v;
while(~scanf("%d",&n)){//dp[i][j]表示以i，j为结尾的两条不相交的路径
for(i=1;i<=n;i++)scanf("%lf %lf",&p[i].x,&p[i].y);
sort(p+1,p+n+1);
for(i=1;i<=n;i++)for(j=1;j<=n;j++)dis[i][j] = Dis(p[i],p[j]), dp[i][j] = inf;
dp[1][1] = 0;
for(i=2;i<=n;i++)
{
for(j = 1;j < i; j++)
{
dp[i][j] = min(dp[i-1][j]+dis[i][i-1], dp[i][j]);//i不与i-1相连，
dp[i][i-1] = min(dp[i-1][j]+dis[j][i], dp[i][i-1]);//i与i-1相连。
}
}
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++)cout< 费用流，把每一个点拆点，中间连流量为1，费用为负无穷的边，代表该点必须选择，两两之间连流量为1，费用为两点距离的边，起点，终点连边，流量为1，费用为0.
代表可以增广两次。
代码：
/* ***********************************************
Author :_rabbit
Created Time :2014/5/17 9:42:51
File Name :6.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include
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