SCU 4488 king's trouble II(dp||枚举)
2017-03-18 18:11
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king's trouble II
Now there is a problem that he worried that he want to choose a largest square of his territory to build a palace.
Can you help him?
For simplicity, we use a matrix to represent the territory as follows:
0 0 0 0 0
0 1 0 1 0
1 1 1 1 0
0 1 1 0 0
0 0 1 0 0
Every single number in the matrix represents a piece of land, which is a 1*1 square
1 represents that this land has been occupied
0 represents not
Obviously the side-length of the largest square is 2
For each case
The first line has two integers N and M representing the length and width of the matrix
Then M lines follows to describe the matrix
1≤N,M≤1000
5 5
0 0 0 0 0
0 1 0 1 0
1 1 0 1 0
0 1 1 0 0
0 0 1 0 0
5 5
0 0 0 0 0
0 1 0 1 0
1 1 1 1 0
0 1 1 0 0
0 0 1 0 0
2
ps:这一题初看的时候真的是没有思路,看了别人的代码才想起可以dp,还可以枚举长度
代码(dp):
代码(枚举):
总结:这一题的做法和那种到终点有多少条路一样,都是求最大或者是最多的,所以可以通过上一步的状态来得到下一步的状态,因此用dp
而枚举我觉得则是这种找最大面积或者是最大边长类型的通解
Description
Long time ago, a king occupied a vast territory.Now there is a problem that he worried that he want to choose a largest square of his territory to build a palace.
Can you help him?
For simplicity, we use a matrix to represent the territory as follows:
0 0 0 0 0
0 1 0 1 0
1 1 1 1 0
0 1 1 0 0
0 0 1 0 0
Every single number in the matrix represents a piece of land, which is a 1*1 square
1 represents that this land has been occupied
0 represents not
Obviously the side-length of the largest square is 2
Input
The first line of the input contains a single integer t (1≤t≤5) — the number of cases.For each case
The first line has two integers N and M representing the length and width of the matrix
Then M lines follows to describe the matrix
1≤N,M≤1000
Output
For each case output the the side-length of the largest squareSample Input
25 5
0 0 0 0 0
0 1 0 1 0
1 1 0 1 0
0 1 1 0 0
0 0 1 0 0
5 5
0 0 0 0 0
0 1 0 1 0
1 1 1 1 0
0 1 1 0 0
0 0 1 0 0
Sample Output
12
ps:这一题初看的时候真的是没有思路,看了别人的代码才想起可以dp,还可以枚举长度
代码(dp):
#include<stdio.h> #include<string.h> #define maxn 1000+10 #define max(a,b) (a>b?a:b) #define min(a,b) (a<b?a:b) int dp[maxn][maxn]; int n,m,ans; int main() { int t; scanf("%d",&t); while(t--) { memset(dp,0,sizeof(dp)); scanf("%d%d",&n,&m); for(int i=0; i<n; i++) for(int j=0; j<m; j++) d500 scanf("%d",&dp[i][j]); ans=0; for(int i=1; i<n; i++) for(int j=1; j<m; j++) { if(dp[i][j]!=0) dp[i][j]=min(dp[i-1][j],min(dp[i][j-1],dp[i-1][j-1]))+1; ans=max(ans,dp[i][j]); } printf("%d\n",ans); } return 0; }
代码(枚举):
#include<stdio.h> #include<string.h> #define maxn 1000+10 #define min(a,b) (a<b?a:b) #define max(a,b) (a>b?a:b) int a[maxn][maxn]; int n,m; int judge(int x,int y,int k) { for(int i=x; i<=x+k; i++)//判断第k列是否符合条件 if(!a[i][y+k]) return 0; for(int i=y; i<=y+k; i++)//判断第k行是否符合条件 if(!a[x+k][i]) return 0; return 1; } int main() { int t; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); for(int i=0; i<n; i++) for(int j=0; j<m; j++) scanf("%d",&a[i][j]); int ans=0,k; for(int i=0; i<n; i++) for(int j=0; j<m; j++) if(a[i][j]) { int ed=min(n-i,m-j);//到最底端还有多长 for(k=1; k<ed; k++)//枚举长度 if(!judge(i,j,k)) break; ans=max(ans,k); } printf("%d\n",ans); } return 0; }
总结:这一题的做法和那种到终点有多少条路一样,都是求最大或者是最多的,所以可以通过上一步的状态来得到下一步的状态,因此用dp
而枚举我觉得则是这种找最大面积或者是最大边长类型的通解
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