usaco16jan radio contact

题目地址

题解

水DP。
按照时间变化递推即可。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55

#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cctype>
#define INF 2000000000
using namespace std;
typedef long long ll;
int (){
int f=1,x=0;char c=getchar();
while(c<'0'||c>'9'){if(c=='-')f=-f;c=getchar();}
while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();
return f*x;
}
int n,m,pos1[1005][2],pos2[1005][2],f[1005][1005];
char s1[1005],s2[1005];
void movement(char c,int &x,int &y){
if(c=='N')y++;if(c=='S')y--;
if(c=='E')x++;if(c=='W')x--;
}
int sqr(int a,int b){
return (pos1[a][0]-pos2[b][0])*(pos1[a][0]-pos2[b][0])+
(pos1[a][1]-pos2[b][1])*(pos1[a][1]-pos2[b][1]);
}
void init(){
n=read(),m=read();
pos1[0][0]=read(),pos1[0][1]=read();
pos2[0][0]=read(),pos2[0][1]=read();
scanf("%s%s",s1+1,s2+1);
for(int i=1,x=pos1[0][0],y=pos1[0][1];i<=n;i++)
movement(s1[i],x,y),
pos1[i][0]=x,pos1[i][1]=y;
for(int i=1,x=pos2[0][0],y=pos2[0][1];i<=m;i++)
movement(s2[i],x,y),
pos2[i][0]=x,pos2[i][1]=y;
}
void solve(){
memset(f,0x14,sizeof(f));
f[0][0]=0;
for(int i=1;i<=n;i++)
f[i][0]=f[i-1][0]+sqr(i,0);
for(int i=1;i<=m;i++)
f[0][i]=f[0][i-1]+sqr(0,i);
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
f[i][j]=min(f[i-1][j],f[i][j-1]),
f[i][j]=min(f[i][j],f[i-1][j-1]),
f[i][j]+=sqr(i,j);
printf("%dn",f[n][m]);
}
int main(){
init();
solve();
return 0;
}