Usaco 17.Jan Copper Set T3

In the morning in the Usaco month of the copper, t1t2 is used to kill, but T3 card a morning, the following to the topic:

Test instructions is probably the input of a n*n matrix, the matrix element only 0 and 12 states, each operation with the top left corner of a matrix in the upper left vertex of a matrix, the matrix of all elements of the state change (that is, 0 to 1, 1 to 0), the matrix of the elements all become 0 the minimum number of times.

The first time to see the sample when thought is a Dfs or BFS search questions, and then decisively wrote the DFS, it is normal to WA. But in fact this problem needs to use greedy ...

As mentioned in the topic, a matrix is changed two times after the original state, that is, if the lower right corner of the point left to the last processing, there is a great possibility to repeat a certain operation, so the bottom right corner of each operation should be from the lower right corner of the search, since this, each operation of the matrix should be how to choose?

Because the final thing is to change all the elements to 0, the optimal solution is to ensure that the bottom-right vertex of each operation is 1, and that the matrix contains the highest number of 1.

Then the strategy is already there. The key code is as follows:

Read the data in advance and ask for prefixes and

for (;;)
{
BOOL Flag=false;
for (int i=1;i<=n;i++)
{
for (int j=1;j<=n;j++)
if (a[i][j]== ' 1 ') {flag=true;break;}
if (flag) break;
}
if (!flag) break;//The judgment matrix is all 0
ans++;
int maxx=0,x=0,y=0;
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)
if (sum[i][j]>maxx&&a[i][j]== ' 1 ')//Find the lower right point of the matrix that has a state of 1 and contains the highest number of elements with a state of 1
{
MAXX=SUM[I][J];
X=i;y=j;
}
for (int i=1;i<=x;i++)
{
for (int j=1;j<=y;j++)
{
if (a[i][j]== ' 1 ') a[i][j]= ' 0 ';
elsea[i][j]= ' 1 ';//change the state of all elements in the selected matrix
}
}
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)
sum[i][j]=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+a[i][j]-' 0 ';//data changes after the matrix prefix and
}

Output answer

Time complexity O (n^4) with data range from 1 to 10, so no timeouts

Usaco 17.Jan Copper Set T3