题意
n辆矿车去两个矿洞,最近3次食物不同的种数为矿产量,有3种食物,问最大矿产量
题解
Dp,f[i][a1][a2][a3][a4] 表示处理到第i辆矿车,1号矿洞和2号矿洞前两次的食物为a1,a2,a3,a4
调试记录
算val时要把当前的放在最前面(因为函数的判断)
#include <cstdio>
#include <algorithm>
#include <cstring>
#define maxn 100005
using namespace std;
int f[2][4][4][4][4];
int n, a[maxn];
int val(int a, int b, int c){
int res = 1;
if (a && a != b && a != c) res++;
if (b && b != c) res++;
return res;
}
int main(){
scanf("%d", &n);
memset(f, -1, sizeof f);
char s[maxn];
scanf("%s", s + 1);
for (int i = 1; i <= n; i++)
if (s[i] == 'M') a[i] = 1;
else if (s[i] == 'B') a[i] = 2;
else if (s[i] == 'F') a[i] = 3;
f[0][0][0][0][0] = 0;
for (int i = 1; i <= n; i++){
for (int pre1 = 0; pre1 < 4; pre1++)
for (int pre2 = 0; pre2 < 4; pre2++)
for (int pre3 = 0; pre3 < 4; pre3++)
for (int pre4 = 0; pre4 < 4; pre4++){
if (f[(i + 1) % 2][pre1][pre2][pre3][pre4] == -1) continue;
f[i % 2][pre2][a[i]][pre3][pre4] = max(f[i % 2][pre2][a[i]][pre3][pre4], f[(i + 1) % 2][pre1][pre2][pre3][pre4] + val(pre1, pre2, a[i]));
f[i % 2][pre1][pre2][pre4][a[i]] = max(f[i % 2][pre1][pre2][pre4][a[i]], f[(i + 1) % 2][pre1][pre2][pre3][pre4] + val(pre3, pre4, a[i]));
}
memset(f[(i + 1) % 2], -1, sizeof(f[(i + 1) % 2]));
}
int ans = 0;
for (int pre1 = 0; pre1 < 4; pre1++)
for (int pre2 = 0; pre2 < 4; pre2++)
for (int pre3 = 0; pre3 < 4; pre3++)
for (int pre4 = 0; pre4 < 4; pre4++)
ans = max(ans, f[n % 2][pre1][pre2][pre3][pre4]);
printf("%d\n", ans);
return 0;
}