P4401 [IOI2007]Miners 矿工配餐 DP
题目链接
大力DP即可.
(f[x][a_1][a_2][b_1][b_2])表示到第(x)天, 第一个矿洞前两天的饭是(a_1, a_2), 第二个矿洞前两天的饭是(b_1, b_2).
转移 : (f[x+1][a2][t][b1][b2] = max(f[x+1][a2][t][b1][b2], f[x][a1][a2][b1][b2] + tmp)),
(f[x+1][a1][a2][b2][t] = max(f[x+1][a1][a2][b2][t], f[x][a1][a2][b1][b2] + tmp));
这两个方程分别表示把第(x + 1)天的饭给了(1)矿洞, (2)矿洞, 然后我们可以根据这三天的饭的情况算出产煤量(tmp).
因为这个题卡空间, 所以第一位要滚掉.
#include <bits/stdc++.h>
using namespace std;
inline long long read() {
long long s = 0, f = 1; char ch;
while(!isdigit(ch = getchar())) (ch == '-') && (f = -f);
for(s = ch ^ 48;isdigit(ch = getchar()); s = (s << 1) + (s << 3) + (ch ^ 48));
return s * f;
}
int n, ans;
char ch[100005];
int f[2][4][4][4][4];
int calc(int x, int y, int z) {
int res = 1;
if(x != 0 && x != y && x != z) res ++;
if(y != 0 && y != z) res ++;
return res;
}
int main() {
n = read(); cin >> ch;
memset(f, -1, sizeof(f)); f[0][0][0][0][0] = 0;
for(int i = 0;i < n; i++) {
int x = i % 2, y = (i + 1) % 2;
for(int a1 = 0;a1 <= 3; a1++) {
for(int a2 = 0;a2 <= 3; a2++) {
for(int b1 = 0;b1 <= 3; b1++) {
for(int b2 = 0;b2 <= 3; b2++) {
if(f[x][a1][a2][b1][b2] == -1) continue ;
int t = ch[i] == 'M' ? 1 : ch[i] == 'B' ? 2 : 3;
int tmp = calc(a1, a2, t);
f[y][a2][t][b1][b2] = max(f[y][a2][t][b1][b2], f[x][a1][a2][b1][b2] + tmp);
tmp = calc(b1, b2, t);
f[y][a1][a2][b2][t] = max(f[y][a1][a2][b2][t], f[x][a1][a2][b1][b2] + tmp);
}
}
}
}
}
for(int a1 = 0;a1 <= 3; a1++) for(int a2 = 0;a2 <= 3; a2++)
for(int b1 = 0;b1 <= 3; b1++) for(int b2 = 0;b2 <= 3; b2++) ans = max(ans, f[n % 2][a1][a2][b1][b2]);
printf("%d", ans);
return 0;
}