洛谷 P3133 [USACO16JAN]Radio Contact G

题目传送门

解题思路:

f[i][j]表示FJ走了i步,Bessie走了j步的最小消耗值.方程比较好推.

横纵坐标要搞清楚,因为这东西WA了半小时.

AC代码:

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 
 5 using namespace std;
 6 
 7 long long n,m,x,y,xx,yy,f[1002][1002];
 8 long long x1[2003],y11[2003],x2[2003],y2[2003];
 9 string l,l1;
10 
11 inline long long min(long long a,long long b) {
12     if(a > b) return b;
13     return a;
14 }
15 
16 inline long long len(int id,int ii) {
17     long long u = 0;
18     long long lenx = abs(x1[id] - x2[ii]);
19     long long leny = abs(y11[id] - y2[ii]);
20     u = lenx * lenx + leny * leny;
21     return u;
22 }
23 
24 int main() {
25     scanf("%lld%lld%lld%lld%lld%lld",&n,&m,&y,&x,&yy,&xx);
26     cin >> l >> l1;
27     x1[0] = x;y11[0] = y;x2[0] = xx;y2[0] = yy;
28     for(int i = 1;i <= n; i++) {
29         if(l[i-1] == 'W') y--;
30         if(l[i-1] == 'E') y++;
31         if(l[i-1] == 'N') x++;
32         if(l[i-1] == 'S') x--;
33         x1[i] = x;y11[i] = y;
34     }
35     for(int i = 1;i <= m; i++) {
36         if(l1[i-1] == 'W') yy--;
37         if(l1[i-1] == 'E') yy++;
38         if(l1[i-1] == 'N') xx++;
39         if(l1[i-1] == 'S') xx--;
40         x2[i] = xx;y2[i] = yy;
41     }
42     for(int i = 1;i <= n; i++) f[i][0] = f[i-1][0] + len(i,0);
43     for(int j = 1;j <= m; j++) f[0][j] = f[0][j-1] + len(0,j);
44     for(int i = 1;i <= n; i++)
45         for(int j = 1;j <= m; j++)
46             f[i][j] = min(f[i-1][j],min(f[i-1][j-1],f[i][j-1])) + len(i,j);
47     printf("%lld",f[n][m]);
48     return 0;
49 }