Wioi 1043 -- number of squares

Description:

Square plot with N * n (n <= 10, we fill in some squares in a positive integer, and other squares in a number 0. As shown in (see example ):

 

A person starts from the point in the upper left corner of the graph and can walk down or right until the B point in the lower right corner. On the way he walked, he could take away the number from the square (the square after the square is removed will become a number 0 ).

This person takes two steps from A.M. To a.m. and tries to find two such paths to maximize the sum of the obtained numbers.

 

Ideas:

(1) DP, 4-dimensional, DP [I1, J1, I2, J2] indicates the maximum value obtained when two paths go to (I1, J1) and (I2, J2) respectively.

 1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5  6 int a[11][11]; 7 int dp[11][11][11][11]; 8 const int INF = 999999999; 9 10 int operDp(int n)11 {12     int i1, j1, i2, j2;13     memset(dp,0,sizeof(dp));14     for(i1 = 1; i1 <= n; i1++)15     for(j1 = 1; j1 <= n; j1++)16     for(i2 = 1; i2 <= n; i2++)17     for(j2 = 1; j2 <= n; j2++)18     {19         int tmp = -INF;20         tmp = max(tmp, dp[i1-1][j1][i2-1][j2]);21         tmp = max(tmp, dp[i1-1][j1][i2][j2-1]);22         tmp = max(tmp, dp[i1][j1-1][i2-1][j2]);23         tmp = max(tmp, dp[i1][j1-1][i2][j2-1]);24         if(i1 == i2 && j1 == j2)25             dp[i1][j1][i2][j2] = tmp + a[i1][j1];26         else27             dp[i1][j1][i2][j2] = tmp + a[i1][j1] + a[i2][j2];28     }29     return dp[n][n][n][n];30 }31 32 int main()33 {34     int n, r, c, v;35     while(scanf("%d",&n) != EOF)36     {37         memset(a,0,sizeof(a));38         while(true)39         {40             scanf("%d%d%d",&r,&c,&v);41             if(!r && !c && !v) break;42             a[r][c] = v;43         }44         printf("%d\n",operDp(n));45     }46     return 0;47 }

(2) reduce the dimension of the DP state

DP [k] [I] [J] indicates that step K is taken. Step I is taken from the first to the right, and step J is taken from the second to the right. The maximum value of K is n, n to the right, that is, 2 * n

 1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5  6 int a[11][11]; 7 int dp[21][11][11]; 8 const int INF = 999999999; 9 10 int operDp(int n)11 {12     int i, j, k;13     memset(dp,0,sizeof(dp));14     for(k = 1; k <= 2 * n; k++)15     for(i = 1; i <= k; i++)16     for(j = 1; j <= k; j++)17     {18         int tmp = -INF;19         tmp = max(tmp, dp[k-1][i-1][j-1]);20         tmp = max(tmp, dp[k-1][i-1][j]);21         tmp = max(tmp, dp[k-1][i][j-1]);22         tmp = max(tmp, dp[k-1][i][j]);23         if(i == j) dp[k][i][j] = tmp + a[k-i+1][i];24         else dp[k][i][j] = tmp + a[k-i+1][i] + a[k-j+1][j];25     }26     return dp[2*n][n][n];27 }28 29 int main()30 {31     int n, r, c, v;32     while(scanf("%d",&n) != EOF)33     {34         memset(a,0,sizeof(a));35         while(true)36         {37             scanf("%d%d%d",&r,&c,&v);38             if(!r && !c && !v) break;39             a[r][c] = v;40         }41         printf("%d\n",operDp(n));42     }43     return 0;44 }