LOJ#2668 书法家

题意：要在一张网格纸上画出NOI图形，使得所占格子的权值和最大。

解：暴力DP即可...

从左往右，每个字母都可以被划分成三块，且每块都可用上下两维来表示。

于是一块一块的DP。考虑如何O(1)转移。显然只有N的中间那一块不好转移，别的都是直接转移。

N的三块的两个连接处之间，可以枚举必须持平的那个端点，另一个用前缀最值。

N第二块内部，考虑枚举后一个矩形的上边界，逐步扩展下边界。此时发现每扩展一步，可能的决策集合会增加：左边一格，下边界为刚扩展的那一格，上边界在枚举的上边界以上的所有状态。

于是对于每个下边界，预处理出上边界从上到下的一个前缀最值，加到决策集合里取max即可。

#include <bits/stdc++.h>

const int INF = 0x3f3f3f3f;

#define g f[FLAG]
#define h f[FLAG ^ 1]

int f[2][510][155][155], FLAG, n, m;
int s[155][510], p[510], temp[155][155];

inline int getSum(int i, int l, int r) {
    if(l > r) return 0;
    return s[r][i] - s[l - 1][i];
}

inline void out() {
    for(int i = 1; i <= m; i++) {
        for(int up = 1; up <= n; up++) {
            for(int down = up; down <= n; down++) {
                printf("%d ", g[i][up][down]);
            }
        }
        puts("");
    }
    puts("");
    return;
}

inline void outp() {
    printf("p : ");
    for(int i = 1; i <= m; i++) {
        printf("%d ", p[i]);
    }
    puts("");
    puts("");
    return;
}

int main() {
    //printf("%d 
", sizeof(f) / 1048576);

    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= m; j++) {
            scanf("%d", &s[i][j]);
            s[i][j] += s[i - 1][j];
        }
    }

    FLAG = 1; /// N 1
    memset(g, ~0x3f, sizeof(g));
    for(int i = 1; i <= m; i++) {
        for(int up = 1; up < n; up++) {
            for(int down = up + 1; down <= n; down++) {
                g[i][up][down] = std::max(0, g[i - 1][up][down]) + getSum(i, up, down);
            }
        }
    }

    //out();

    FLAG ^= 1; /// N 1.5
    memset(g, ~0x3f, sizeof(g));
    for(int i = 1; i <= m; i++) {
        for(int up = 1; up < n; up++) {
            int large = -INF;
            for(int down = n - 1; down >= up; down--) {
                large = std::max(large, h[i - 1][up][down + 1]);
                g[i][up][down] = large + getSum(i, up, down);
            }
        }
    }

    //out();

    FLAG ^= 1; /// N 2
    //memset(g, ~0x3f, sizeof(g));
    memcpy(g, h, sizeof(h));
    for(int i = 1; i <= m; i++) {
        memset(temp, ~0x3f, sizeof(temp));
        for(int down = 1; down <= n; down++) {
            for(int up = 1; up <= down; up++) {
                temp[down][up] = std::max(temp[down][up - 1], std::max(g[i - 1][up][down], h[i - 1][up][down]));
            }
        }
        for(int up = 1; up <= n; up++) {
            int large = temp[up - 1][up - 1];
            for(int down = up; down <= n; down++) {
                large = std::max(large, temp[down][up]);
                g[i][up][down] = std::max(large + getSum(i, up, down), h[i][up][down]);
            }
        }
    }

    //out();

    FLAG ^= 1; /// N 2.5
    memset(g, ~0x3f, sizeof(g));
    for(int i = 1; i <= m; i++) {
        for(int down = 2; down <= n; down++) {
            int large = -INF;
            for(int up = down - 1; up >= 1; up--) {
                large = std::max(large, h[i - 1][up + 1][down]);
                g[i][up][down] = large + getSum(i, up, down);
            }
        }
    }

    //out();

    FLAG ^= 1; /// N 3
    memset(g, ~0x3f, sizeof(g));
    for(int i = 1; i <= m; i++) {
        for(int up = 1; up < n; up++) {
            for(int down = up + 1; down <= n; down++) {
                g[i][up][down] = std::max(h[i][up][down], g[i - 1][up][down] + getSum(i, up, down));
            }
        }
    }

    //out();

    /// get p : max of g
    memset(p, ~0x3f, sizeof(p));
    for(int i = 1; i <= m; i++) {
        p[i] = p[i - 1];
        for(int up = 1; up <= n; up++) {
            for(int down = 1; down <= n; down++) {
                p[i] = std::max(p[i], g[i][up][down]);
            }
        }
    }

    //outp();

    //out();

    //printf(" O 1 
");

    FLAG ^= 1; /// O 1
    memset(g, ~0x3f, sizeof(g));
    for(int i = 5; i <= m; i++) {
        for(int up = 1; up < n - 1; up++) {
            for(int down = up + 2; down <= n; down++) {
                g[i][up][down] = p[i - 2] + getSum(i, up, down);
            }
        }
    }

    //out();

    //printf(" O 2 
");

    FLAG ^= 1; /// O 2
    memset(g, ~0x3f, sizeof(g));
    for(int i = 1; i <= m; i++) {
        for(int up = 1; up < n - 1; up++) {
            for(int down = up + 2; down <= n; down++) {
                g[i][up][down] = std::max(h[i - 1][up][down], g[i - 1][up][down]) + getSum(i, up, up) + getSum(i, down, down);
            }
        }
    }

    //out();

    FLAG ^= 1; /// O 3
    memset(g, ~0x3f, sizeof(g));
    for(int i = 1; i <= m; i++) {
        for(int up = 1; up < n - 1; up++) {
            for(int down = up + 2; down <= n; down++) {
                g[i][up][down] = h[i - 1][up][down] + getSum(i, up, down);
            }
        }
    }

    //out();

    /// get p : max of g
    memset(p, ~0x3f, sizeof(p));
    for(int i = 1; i <= m; i++) {
        p[i] = p[i - 1];
        for(int up = 1; up < n - 1; up++) {
            for(int down = up + 2; down <= n; down++) {
                p[i] = std::max(p[i], g[i][up][down]);
            }
        }
    }

    //outp();

    //out();

    FLAG ^= 1; /// I 1
    memset(g, ~0x3f, sizeof(g));
    for(int i = 5; i <= m; i++) {
        for(int up = 1; up < n - 1; up++) {
            for(int down = up + 2; down <= n; down++) {
                g[i][up][down] = std::max(p[i - 2], g[i - 1][up][down]) + getSum(i, up, up) + getSum(i, down, down);
            }
        }
    }

    //out();

    FLAG ^= 1; /// I 2
    memset(g, ~0x3f, sizeof(g));
    for(int i = 1; i <= m; i++) {
        for(int up = 1; up < n - 1; up++) {
            for(int down = up + 2; down <= n; down++) {
                g[i][up][down] = std::max(g[i - 1][up][down], h[i - 1][up][down]) + getSum(i, up, down);
            }
        }
    }

    //out();
    int ans = -INF;

    FLAG ^= 1; /// I 3
    memset(g, ~0x3f, sizeof(g));
    for(int i = 1; i <= m; i++) {
        for(int up = 1; up < n - 1; up++) {
            for(int down = up + 2; down <= n; down++) {
                g[i][up][down] = std::max(g[i - 1][up][down], h[i - 1][up][down]) + getSum(i, up, up) + getSum(i, down, down);
                ans = std::max(g[i][up][down], ans);
            }
        }
    }

    //out();

    printf("%d
", ans);
    return 0;
}