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.

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如果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 <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
#include <queue>
#include <set>
#include <algorithm>
#include <stdlib.h>
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<<dp[i][j]<<" ";
			cout<<endl;
		}
		printf("%.2lf
",dp[n][n-1]+dis[n][n-1]);
    }
    return 0;
}

费用流。把每个点拆点，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.

代表能够增广两次。

代码：

/* ***********************************************
Author :_rabbit
Created Time :2014/5/17 9:42:51
File Name :6.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 1001000
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
const int maxn=4010;
const int maxm=200000;
struct Edge{
    int next,to,cap;
	double cost;
    Edge(int _next=0,int _to=0,int _cap=0,double _cost=0){
        next=_next;to=_to;cap=_cap;cost=_cost;
    }
}edge[maxm];
int head[maxn],vis[maxn],pre[maxn],n,tol;
double dis[maxn];
void addedge(int u,int v,int cap,double cost){
    edge[tol]=Edge(head[u],v,cap,cost);head[u]=tol++;
    edge[tol]=Edge(head[v],u,0,-cost);head[v]=tol++;
}
bool spfa(int s,int t){
    queue<int> q;
    for(int i=0;i<=n;i++)
        dis[i]=INF,vis[i]=0,pre[i]=-1;
    dis[s]=0;vis[s]=1;q.push(s);
    while(!q.empty()){
        int u=q.front();q.pop();vis[u]=0;
	//	cout<<"u="<<u<<" "<<dis[u]<<endl;
        for(int i=head[u];i!=-1;i=edge[i].next){
            int v=edge[i].to;
            if(edge[i].cap>0&&dis[v]>dis[u]+edge[i].cost){
                dis[v]=dis[u]+edge[i].cost;
                pre[v]=i;
                if(!vis[v])vis[v]=1,q.push(v);
            }
        }
    }
    if(pre[t]==-1)return 0;
    return 1;
}
void fun(int s,int t,int &flow,double &cost){
    flow=0;cost=0;
    while(spfa(s,t)){
        int MIN=INF;
        for(int i=pre[t];i!=-1;i=pre[edge[i^1].to])
            if(MIN>edge[i].cap)MIN=edge[i].cap;
        for(int i=pre[t];i!=-1;i=pre[edge[i^1].to])
            edge[i].cap-=MIN,edge[i^1].cap+=MIN,cost+=edge[i].cost*MIN;
        flow+=MIN;
    }
}
struct Point{
	double x,y;
}pp[10000];
double dist(Point a,Point b){
	double ss=a.x-b.x;
	double tt=a.y-b.y;
	return sqrt(ss*ss+tt*tt);
}
int main(){
	int m;
//	freopen("data.out","w",stdout);
	while(cin>>m){
		memset(head,-1,sizeof(head));tol=0;
		for(int i=1;i<=m;i++)cin>>pp[i].x>>pp[i].y;
		for(int i=1;i<=m;i++){
			for(int j=i+1;j<=m;j++){
				double dd=dist(pp[i],pp[j]);
				addedge(i+m,j,1,dd);
			}
			addedge(i,i+m,1,-INF);
		}
		addedge(1,m+1,1,0);
		addedge(m,2*m,1,0);
		n=2*m;
		int flow;double cost;
		fun(1,2*m,flow,cost);
		printf("%.2lf
",cost+m*INF);
	}
	return 0;
}