Bzoj 1088: [SCOI2005]掃雷Mine (DP)

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Bzoj 1088: [SCOI2005]掃雷Mine

怒寫一發,算不上DP的遊戲題
知道了前\(i-1\)項,第\(i\)項會被第二列的第\(i-1\)得知
設\(f[i]\)為第一列的第\(i\)行位置是否有雷,有雷的話,\(f[i] = 1\),無雷\(f[i] = 0\)
\(a[i]\)就是題目讀入的東西.
那麼轉移方程就是\(f[i] = a[i - 1] - f[i - 1] - f[i - 2]\)
不滿足限制的時候就是\(f[i] < 0\) 或者$ f[i] > 1$
第一個位置討論一下即可.進行上面的遞推.

#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;}

Bzoj 1088: [SCOI2005]掃雷Mine (DP)