HDU 2224 The shortest path
The shortest path
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1507 Accepted Submission(s): 773
Before you reach the rightmost point Pn, you can only visit the points those have the bigger x-coordinate value. For example, you are at Pi now, then you can only visit Pj(j > i). When you reach Pn, the rule is changed, from now on you can only visit the points those have the smaller x-coordinate value than the point you are in now, for example, you are at Pi now, then you can only visit Pj(j < i). And in the end you back to P1 and the tour is over.
You should visit all points in this tour and you can visit every point only once.
1 1
2 3
3 1
Hint: The way 1 - 3 - 2 - 1 makes the shortest path.
#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n;
double dis[][],f[][];
struct nond{
int x,y;
}v[];
int cmp(nond a,nond b){
if(a.x==b.x) return a.y<b.y;
return a.x<b.x;
}
void pre(){
for(int i=;i<=n;i++)
for(int j=i;j<=n;j++)
dis[i][j]=dis[j][i]=sqrt((double)(v[i].x-v[j].x)*(v[i].x-v[j].x)+(v[i].y-v[j].y)*(v[i].y-v[j].y));
}
int main(){
while(scanf("%d",&n)!=EOF){
memset(dis,,sizeof(dis));
for(int i=;i<=n;i++)
scanf("%d%d",&v[i].x,&v[i].y);
sort(v+,v++n,cmp);
pre();
f[][]=f[][]=dis[][];
f[][]=*dis[][];
for(int i=;i<=n;i++){
for(int j=;j<i-;j++)
f[i][j]=f[j][i]=f[i-][j]+dis[i][i-];
f[i][i-]=f[i-][i]=f[i][i]=0x7f7f7f7f;
for(int j=;j<=i-;j++)
f[i-][i]=f[i][i-]=min(f[i][i-],f[j][i-]+dis[j][i]);
for(int j=;j<=i;j++)
f[i][i]=min(f[i][i],f[j][i]+dis[j][i]);
}
printf("%.2lf\n",f[n][n]);
}
}
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