Bzoj 1088: [SCOI2005]扫雷Mine (DP)

Bzoj 1088: [SCOI2005]扫雷Mine

怒写一发,算不上DP的游戏题
知道了前\(i-1\)项,第\(i\)项会被第二列的第\(i-1\)得知
设\(f[i]\)为第一列的第\(i\)行位置是否有雷,有雷的话,\(f[i] = 1\),无雷\(f[i] = 0\)
\(a[i]\)就是题目读入的东西.
那么转移方程就是\(f[i] = a[i - 1] - f[i - 1] - f[i - 2]\)
不满足限制的时候就是\(f[i] < 0\) 或者$ f[i] > 1$
第一个位置讨论一下即可.进行上面的递推.

#include <iostream>
#include <cstdio>
const int maxN = 10000 + 7;
 
int f[maxN],ans,a[maxN];
 
inline int read() {
    int x = 0,f = 1;char c = getchar();
    while(c < '0' || c > '9') {if(c == '-')f = -1;c = getchar();}
    while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}
    return x * f;
}
 
int n;
void work() {
    for(int i = 2;i <= n;++ i) {
        f[i] = a[i - 1] - f[i - 1] - f[i - 2];
        if(f[i] < 0 || f[i] > 1) return ;
    }
    if(a[n] != f[n] + f[n - 1])return ;
    ans ++;
    return ;
}
 
int main() {
    n = read();
    for(int i = 1;i <= n;++ i) 
        a[i] = read();
    for(int i = 0;i < 2;++ i) 
        f[1] = i,work();
    printf("%d\n", ans);
    return 0;
}