标签：event == san while display ret 处理 set col
$Description$
求最大子矩阵.$1<=N<=100$
$Sol$
这个数据范围暴力就可以过了$qwq$
但还是要讲一下优化:预处理前缀和$sum[i][j]$表示的是$\sum_{k=1}^{i}a[k][j]$.然后就只要枚举矩阵的上下两边,于是求最大子矩阵转化成看求最大子段和.
注意矩阵里的元素可以为负!
$Code$
#include<iostream> #include<cstdio> #define il inline #define Rg register #define go(i,a,b) for(Rg int i=a;i<=b;++i) #define yes(i,a,b) for(Rg int i=a;i>=b;--i) #define mem(a,b) memset(a,b,sizeof(a)) #define ll long long #define db double using namespace std; il int read() { Rg int x=0,y=1;char c=getchar(); while(c<‘0‘||c>‘9‘){if(c==‘-‘)y=-1;c=getchar();} while(c>=‘0‘&&c<=‘9‘){x=(x<<1)+(x<<3)+c-‘0‘;c=getchar();} return x*y; } int n,as,a[110][110]; int main() { n=read(); go(i,1,n)go(j,1,n)a[i][j]=a[i][j-1]+a[i-1][j]-a[i-1][j-1]+read(); go(x1,1,n)go(y1,1,n) go(x2,x1,n)go(y2,y1,n) as=max(as,a[x2][y2]-a[x1-1][y2]-a[x2][y1-1]+a[x1-1][y1-1]); printf("%d\n",as); return 0; }
#include<iostream> #include<cstdio> #define il inline #define Rg register #define go(i,a,b) for(Rg int i=a;i<=b;++i) #define yes(i,a,b) for(Rg int i=a;i>=b;--i) #define mem(a,b) memset(a,b,sizeof(a)) #define ll long long #define db double #define inf 2147483647 using namespace std; il int read() { Rg int x=0,y=1;char c=getchar(); while(c<‘0‘||c>‘9‘){if(c==‘-‘)y=-1;c=getchar();} while(c>=‘0‘&&c<=‘9‘){x=(x<<1)+(x<<3)+c-‘0‘;c=getchar();} return x*y; } int n,as=-inf,a[110][110]; int main() { n=read(); go(i,1,n)go(j,1,n)a[i][j]=a[i-1][j]+read(); go(i,1,n)go(j,i,n) { Rg int nw=0; go(k,1,n) { Rg int qvq=a[j][k]-a[i-1][k]; if(nw>=0)nw+=qvq; else nw=qvq; as=max(as,nw); } } printf("%d\n",as); return 0; }
标签：event == san while display ret 处理 set col
原文地址：https://www.cnblogs.com/forward777/p/11381354.html
