題目鏈接

bzoj3106: [cqoi2013]棋盤遊戲

題解

n = 20
顯然當A不能一步吃掉B時,A必輸
那麽就是stepA - stepB
A想步數多，B想步數少
對抗搜索
可以證明B一定能在n * 4步以內獲勝
記憶化一下

代碼

/*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\n");   
    else printf("BLACK %d\n",dfs(0,0,a,b,c,d)); 
    return 0; 
}

bzoj3106: [cqoi2013]棋盤遊戲