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:

 #include<iostream>
 #include<cstdio>
 #include<cmath>

 using namespace std;

 long long n,m,x,y,xx,yy,f[][];
 long long x1[],y11[],x2[],y2[];
 string l,l1;

 inline long long min(long long a,long long b) {
     if(a > b) return b;
     return a;
 }

 inline long long len(int id,int ii) {
     long long u = ;
     long long lenx = abs(x1[id] - x2[ii]);
     long long leny = abs(y11[id] - y2[ii]);
     u = lenx * lenx + leny * leny;
     return u;
 }

 int main() {
     scanf("%lld%lld%lld%lld%lld%lld",&n,&m,&y,&x,&yy,&xx);
     cin >> l >> l1;
     x1[] = x;y11[] = y;x2[] = xx;y2[] = yy;
     for(int i = ;i <= n; i++) {
         if(l[i-] == 'W') y--;
         if(l[i-] == 'E') y++;
         if(l[i-] == 'N') x++;
         if(l[i-] == 'S') x--;
         x1[i] = x;y11[i] = y;
     }
     for(int i = ;i <= m; i++) {
         if(l1[i-] == 'W') yy--;
         if(l1[i-] == 'E') yy++;
         if(l1[i-] == 'N') xx++;
         if(l1[i-] == 'S') xx--;
         x2[i] = xx;y2[i] = yy;
     }
     for(int i = ;i <= n; i++) f[i][] = f[i-][] + len(i,);
     for(int j = ;j <= m; j++) f[][j] = f[][j-] + len(,j);
     for(int i = ;i <= n; i++)
         for(int j = ;j <= m; j++)
             f[i][j] = min(f[i-][j],min(f[i-][j-],f[i][j-])) + len(i,j);
     printf("%lld",f[n][m]);
     return ;
 }