标签：style blog color os io for div amp
动态规划
题意可以理解为两个人同时从最左点出发，沿着两条不同的路径走到最右点（除了起点和终点每个点走且仅走一次）
状态 dp[i][j]指当前两人分别走到i，j点。且设i>j；
则有：dp[i+1][i]=min (dp[i+1][i],dp[i][j]+dist[i][i+1]);
dp[i+1][j]=min (dp[i+1][j],dp[i][j]+dist[j][i+1]);
因为每个点都要走到，所以每次走到没走到的第一个点。
最终结果是 dp[n-1][i]+dist[i][n]+dist[n-1][n];(0<i<n-1) 最终点要走两次。
1 #include <iostream> 2 #include <algorithm> 3 #include <cstring> 4 #include <cstdio> 5 #include <cmath> 6 using namespace std; 7 8 const double inf=99999999.0; 9 int n; 10 double a[1005][2]; 11 double dist[1005][1005]; 12 double dp[1005][1005]; 13 14 double getdist (double x1,double y1,double x2,double y2){ 15 double x,y; 16 x=fabs (x1-x2); 17 y=fabs (y1-y2); 18 return sqrt (x*x+y*y); 19 } 20 21 int main (){//cout<<inf<<endl; 22 while (~scanf ("%d",&n)){ 23 for (int i=0;i<n;i++){ 24 scanf ("%lf%lf",&a[i][0],&a[i][1]); 25 } 26 for (int i=0;i<n;i++) 27 for (int j=0;j<n;j++) 28 dist[i][j]=getdist (a[i][0],a[i][1],a[j][0],a[j][1]);//cout<<i<<" "<<j<<":"<<dist[i][j]<<" "; 29 for (int i=0;i<=n;i++) 30 for (int j=0;j<=n;j++) 31 dp[i][j]=inf; 32 dp[0][0]=0; 33 for (int i=0;i<n-2;i++){ 34 for (int j=0;(i==0&&j==0)||j<i;j++){ 35 dp[i+1][j]=min (dp[i+1][j],dp[i][j]+dist[i][i+1]); 36 dp[i+1][i]=min (dp[i+1][i],dp[i][j]+dist[j][i+1]);//cout<<dp[i+1][j]<<" "<<dp[i+1][i]<<endl; 37 } 38 }//cout<<endl; 39 double ans=inf; 40 for (int i=0;i<n-2;i++) 41 ans=min (ans,dp[n-2][i]+dist[n-1][n-2]+dist[n-1][i]);//cout<<n-2<<" "<<i<<endl,cout<<dp[n-2][i]+dist[n-1][n-2]+dist[n-1][i]<<endl; 42 printf ("%.2f\n",ans); 43 } 44 return 0; 45 }
标签：style blog color os io for div amp
原文地址：http://www.cnblogs.com/gfc-g/p/3877247.html
