[luoguP1136] 迎接儀式(DP)
阿新 • • 發佈:2017-08-06
需要 i++ amp fin set sca std print ems
傳送門
每個字母只有兩種選擇,變成另一個或者不變。
所以f[i][j][k]表示前i個字母有j個j變成z,有k個z變成j
只需要比較j==k時的答案就行
#include <cstdio> #include <cstring> #define N 505 #define max(x, y) ((x) > (y) ? (x) : (y)) int n, m, ans; char s[N]; int f[N][105][105]; int main() { int i, j, k; scanf("%d %d", &n, &m); scanf("%s", s + 1); memset(f, -0x3f, sizeof(f)); f[0][0][0] = f[1][0][0] = 0; s[1] == ‘j‘ ? f[1][1][0] = 0 : f[1][0][1] = 0; for(i = 2; i <= n; i++) for(j = 0; j <= m; j++) for(k = 0; k <= m; k++) { f[i][j][k] = f[i - 1][j][k]; if(s[i - 1] == ‘j‘ && s[i] == ‘z‘) f[i][j][k] = max(f[i][j][k], f[i - 2][j][k] + 1); if(s[i - 1] == ‘j‘ && s[i] == ‘j‘ && j) f[i][j][k] = max(f[i][j][k], f[i - 2][j - 1][k] + 1); if(s[i - 1] == ‘z‘ && s[i] == ‘z‘ && k) f[i][j][k] = max(f[i][j][k], f[i - 2][j][k - 1] + 1); if(s[i - 1] == ‘z‘ && s[i] == ‘j‘ && j && k) f[i][j][k] = max(f[i][j][k], f[i - 2][j - 1][k - 1] + 1); if(j == k) ans = max(ans, f[i][j][k]); } printf("%d\n", ans); return 0; }
[luoguP1136] 迎接儀式(DP)