[SCOI2009]粉刷匠

Description

Luogu4158

BZOJ1296

Solution

考虑最基础的DP，枚举每个位置和从这个位置向前涂多远，(O(n^5))DP，吸氧过了……

标解是(O(n^4))

Code

#include <algorithm>
#include <bitset>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <map>
#include <queue>
#include <set>
#include <vector>

namespace wyx {

typedef long long ll;
typedef double ld;

ll read() {
    char c;
    ll ans = 0, fl = 1;
    for (c = getchar(); c > '9' || c < '0'; c = getchar())
        if (c == '-') fl = -1;
    ans = c - '0';
    for (c = getchar(); c >= '0' && c <= '9'; c = getchar())
        ans = ans * 10 + c - '0';
    // scanf("%lld
", &ans);
    return fl * ans;
}

const int N = 51;
const int M = 2501;

int f[N][N][M], n, m, T, s[N][N], p[N][N][N];
char a[N][N];
inline int max(const int &x, const int &y) { return x > y ? x : y; }
// f[i][j][t] = f[i][j-k][t-1] + w[j - k, j];
inline int w(const int &i, const int &j, const int &k) {
    return ~p[i][j][k] ? p[i][j][k]
                       : p[i][j][k] = max(s[i][j] - s[i][k - 1],
                                          j - k + 1 - s[i][j] + s[i][k - 1]);
}

void main() {
    n = read();
    m = read();
    T = read();
    int ans = 0;
    memset(p, -1, sizeof p);
    for (int i = 1; i <= n; ++i) {
        scanf("%s", a[i] + 1);
        for (int j = 1; j <= m; ++j) {
            s[i][j] = s[i][j - 1] + a[i][j] - '0';
            for (int t = 0; t <= T; ++t) f[i][j][t] = f[i - 1][m][t];
        }
        for (int t = 0; t <= T; ++t) f[i][0][t] = f[i - 1][m][t];
        for (int j = 1; j <= m; ++j) {
            for (int t = 1; t <= T; ++t) {
                for (int k = j; k > 0; --k) {
                    f[i][j][t] =
                        max(f[i][j][t], f[i][k - 1][t - 1] + w(i, j, k));
                    // ans = max(ans, f[i][j][t]);
                }
            }
        }
        for (int j = 1; j <= m; ++j) ans = max(ans, f[i][j][T]);
    }
    printf("%d
", ans);
}

}  // namespace wyx

int main() {
    wyx::main();
    return 0;
}