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[][][][], FLAG, n, m;
 int s[][], p[], temp[][];

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

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

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

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

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

     //out();

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

     //out();

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

     //out();

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

     //out();

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

     //outp();

     //out();

     //printf(" O 1 \n");

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

     //out();

     //printf(" O 2 \n");

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

     //out();

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

     //out();

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

     //outp();

     //out();

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

     //out();

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

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

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

     //out();

     printf("%d\n", ans);
     return ;
 }

AC代码