题目链接
题解
n = 20
显然当A不能一步吃掉B时,A必输
那么就是stepA - stepB
A想步数多，B想步数少
对抗搜索
可以证明B一定能在n * 4步以内获胜
记忆化一下
代码
/*bzoj 3106*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#define INF 0x3f3f3f3f
inline int read() {
int x = 0,f = 1;
char c = getchar();
while(c < '0' || c > '9') {if(c == '-')f = -1;c = getchar();}
while(c <= '9' && c >= '0') x = x * 10 + c - '0', c = getchar();
return x * f;
}
const int maxn = 21;
int n;
int s[2][70][maxn][maxn][maxn][maxn];
int fs1[5] = {1,0,-1,0,1};
int fs2[5] = {2,0,-2,0,2};
inline bool judge(int x,int y) { return (x >= 1 && x <= n && y >= 1 && y <= n) ? true : false;}
int dfs(int nxt,int step,int a,int b,int c,int d) {
if(step > n * 3) return INF;
if(a == c && b == d) return nxt ? INF : 0;
if(s[nxt][step][a][b][c][d]) return s[nxt][step][a][b][c][d];
int ret = nxt ? INF : 0;
if(nxt) {
for(int i = 0;i < 4;++ i) {
int tx = c + fs1[i],ty = d + fs1[i + 1];
if(judge(tx,ty)) ret = std::min(ret,dfs(nxt ^ 1,step + 1,a,b,tx,ty)) ;
tx = c + fs2[i],ty = d + fs2[i + 1];
if(judge(tx,ty)) ret = std::min(ret,dfs(nxt ^ 1,step + 1,a,b,tx,ty)) ;
}
}
else
for(int i = 0;i < 4;++ i) {
int tx = a + fs1[i],ty = b + fs1[i + 1];
if(judge(tx,ty)) ret = std::max(ret,dfs(nxt ^ 1,step + 1,tx,ty,c,d));
}
return s[nxt][step][a][b][c][d] = ++ret;
}
int main() {
int a,b,c,d;
n = read(),a = read(),b = read(),c = read(),d = read();
if(abs(a - c) + abs(b - d) == 1) printf("WHITE 1
");
else printf("BLACK %d
",dfs(0,0,a,b,c,d));
return 0;
}