UVA1347 旅游(二维递归DP)
旅游
【题目链接】旅游
【题目类型】DP
&题解:
紫书P269 代码很简单,但思路很难。很难能想到要把一个圈分成2条线段,很难想到d(i,j)表示的是已经走过max(i,j)还需要的距离值,当然设d为还需要的距离值,这很常见。
还有也很难想到下一步只能走到i+1。
【时间复杂度】O(n^2)
&代码:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
#define cle(a,val) memset(a,(val),sizeof(a))
#define SI(N) scanf("%d",&(N))
#define SII(N,M) scanf("%d %d",&(N),&(M))
#define SIII(N,M,K) scanf("%d %d %d",&(N),&(M),&(K))
#define rep(i,b) for(int i=0;i<(b);i++)
#define rez(i,a,b) for(int i=(a);i<=(b);i++)
#define red(i,a,b) for(int i=(a);i>=(b);i--)
const ll LINF = 0x3f3f3f3f3f3f3f3f;
#define PU(x) puts(#x);
#define PI(A) cout<<(A)<<endl;
#define DG(x) cout<<#x<<"="<<(x)<<endl;
#define DGG(x,y) cout<<#x<<"="<<(x)<<" "<<#y<<"="<<(y)<<endl;
#define DGGG(x,y,z) cout<<#x<<"="<<(x)<<" "<<#y<<"="<<(y)<<" "<<#z<<"="<<(z)<<endl;
#define PIar(a,n) rep(i,n)cout<<a[i]<<" ";cout<<endl;
#define PIarr(a,n,m) rep(aa,n){rep(bb, m)cout<<a[aa][bb]<<" ";cout<<endl;}
const double EPS = 1e-9 ;
/* //////////////////////// C o d i n g S p a c e //////////////////////// */
const int MAXN = 1000 + 9 ;
struct pnt{
int x,y;
}p[MAXN];
int n;
double d[MAXN][MAXN];
double dis(int i,int j){
return hypot(p[i].x-p[j].x,p[i].y-p[j].y);
}
double dp(int i,int j){
double& ans=d[i][j];
if (i==n-1) return ans=dis(i,n)+dis(j,n);
if (ans>0) return ans;
ans=min(dp(i+1,j)+dis(i,i+1),dp(i+1,i)+dis(j,i+1));
return ans;
}
void Solve()
{
while(~SI(n)){
// dp problem input must start with 1
rez(i,1,n) SII(p[i].x,p[i].y);
cle(d,0);
double ans=dp(1,2)+dis(1,2);
printf("%.2lf\n", ans);
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("1.in", "r", stdin);
freopen("1.out","w",stdout);
#endif
//iostream::sync_with_stdio(false);
//cin.tie(0), cout.tie(0);
// int T;cin>>T;while(T--)
Solve();
return 0;
}