POJ 2677 旅行商问题 双调dp或者费用流

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Description

Input

Output

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 <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\n",dp[n][n-1]+dis[n][n-1]);
    }
    return 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 <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\n",cost+m*INF);
	}
	return 0;
}

POJ 2677 旅行商问题 双调dp或者费用流,布布扣,bubuko.com

POJ 2677 旅行商问题 双调dp或者费用流

标签：des   blog   class   code   c   tar   

原文地址：http://blog.csdn.net/xianxingwuguan1/article/details/26089461