USACO 5.3.4 Big Barn 巨大的牛棚 bigbrn
动态规划。前面的章节中还有一个与这个基本上一样的题。
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状态 o F[i][j] 表示以(i,j)为左上角最大正方形的边长
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初始条件 o 如果(i,N)没有障碍 F[i][N]=1 否则 F[i][N]=0 o 如果(N,i)没有障碍 F[N][i]=1 否则 F[N][i]=0
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状态转移方程 o F[i][j]=min(F[i+1][j],F[i][j+1],F[i+1][j+1])+1;
USER: CmYkRgB123 CmYkRgB123 [cmykrgb1] TASK: bigbrn LANG: C++ Compiling... Compile: OK Executing... Test 1: TEST OK [0.011 secs, 6752 KB] Test 2: TEST OK [0.000 secs, 6752 KB] Test 3: TEST OK [0.000 secs, 6756 KB] Test 4: TEST OK [0.000 secs, 6756 KB] Test 5: TEST OK [0.000 secs, 6756 KB] Test 6: TEST OK [0.000 secs, 6752 KB] Test 7: TEST OK [0.000 secs, 6752 KB] Test 8: TEST OK [0.011 secs, 6756 KB] Test 9: TEST OK [0.011 secs, 6756 KB] Test 10: TEST OK [0.032 secs, 6752 KB] Test 11: TEST OK [0.032 secs, 6752 KB] Test 12: TEST OK [0.032 secs, 6756 KB] Test 13: TEST OK [0.022 secs, 6756 KB] Test 14: TEST OK [0.022 secs, 6752 KB] Test 15: TEST OK [0.043 secs, 6752 KB] All tests OK. Your program ('bigbrn') produced all correct answers! This is your submission #3 for this problem. Congratulations!
/*
ID: cmykrgb1
PROG: bigbrn
LANG: C++
*/
#include <iostream>
#include <fstream>
#define MAXN 1001
using namespace std;
ifstream fi("bigbrn.in");
ofstream fo("bigbrn.out");
int M[MAXN][MAXN],N,ans;
void init()
{
int T,x,y,i;
fi >> N >> T;
for (i=1;i<=T;i++)
{
fi >> x >> y;
M[x][y]=-1;
}
}
inline int min(int a,int b,int c)
{
if (a<=b && a<=c) return a;
if (b<=a && b<=c) return b;
return c;
}
void dynamic()
{
int i,j;
for (i=1;i<=N;i++)
{
if (M[N][i]==0) M[N][i]=1;
else M[N][i]=0;
if (M[i][N]==0) M[i][N]=1;
else M[i][N]=0;
}
for (i=N-1;i>=1;i--)
for (j=N-1;j>=1;j--)
if (M[i][j]==-1)
M[i][j]=0;
else
{
M[i][j]=min(M[i+1][j],M[i][j+1],M[i+1][j+1])+1;
if (M[i][j]>ans)
ans=M[i][j];
}
}
void print()
{
fo << ans << endl;
fi.close();
fo.close();
}
int main()
{
int i,j;
init();
dynamic();
print();
return 0;
}
上次修改时间 2017-05-22