• luogu2331 [SCOI2005]最大子矩阵


    题目大意

      这里有一个n*m的矩阵,请你选出其中k个子矩阵,使得这个k个子矩阵分值之和最大。注意:选出的k个子矩阵不能相互重叠。1≤n≤100,1≤m≤2,1≤k≤10。

    思路

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    
    #define Update(x, y) x = max(x, y)
    const int MAX_ROW = 110, MAX_COL = 5, MAX_K = 15, MAX_STATE = 5, MINF = 0xcfcfcfcf;
    int F[MAX_ROW][MAX_STATE][MAX_K];
    int A[MAX_ROW][MAX_COL];
    int TotRow, TotCol, K;
    
    int DP1()
    {
    	memset(F, 0xcf, sizeof(F));
    	F[0][0][0] = 0;
    	for (int i = 1; i <= TotRow; i++)
    	{
    		for (int k = 0; k <= K; k++)
    		{
    			Update(F[i][0][k], F[i - 1][0][k]),
    			Update(F[i][0][k], F[i - 1][1][k]);
    			
    			if (k >= 1)
    				Update(F[i][1][k], A[i][1] + F[i - 1][0][k - 1]),
    				Update(F[i][1][k], A[i][1] + F[i - 1][1][k - 1]);
    			Update(F[i][1][k], A[i][1] + F[i - 1][1][k]);
    		}
    	}
    	return max(F[TotRow][0][K], F[TotRow][1][K]);
    }
    
    int DP2()
    {
    	memset(F, 0xcf, sizeof(F));
    	F[0][0][0] = 0;
    	for (int i = 1; i <= TotRow; i++)
    		for (int k = 0; k <= K; k++)
    		{
    			for (int j = 0; j <= 4; j++)
    				Update(F[i][0][k], F[i - 1][j][k]);
    
    			if (k >= 1)
    				for (int j = 0; j <= 4; j++)
    					Update(F[i][1][k], A[i][1] + F[i - 1][j][k - 1]);
    			Update(F[i][1][k], A[i][1] + F[i - 1][1][k]);
    			Update(F[i][1][k], A[i][1] + F[i - 1][3][k]);
    
    			if (k >= 1)
    				for (int j = 0; j <= 4; j++)
    					Update(F[i][2][k], A[i][2] + F[i - 1][j][k - 1]);
    			Update(F[i][2][k], A[i][2] + F[i - 1][2][k]);
    			Update(F[i][2][k], A[i][2] + F[i - 1][3][k]);
    
    			if (k >= 2)
    				for (int j = 0; j <= 4; j++)
    					Update(F[i][3][k], A[i][1] + A[i][2] + F[i - 1][j][k - 2]);
    			if (k >= 1)
    				Update(F[i][3][k], A[i][1] + A[i][2] + F[i - 1][1][k - 1]),
    				Update(F[i][3][k], A[i][1] + A[i][2] + F[i - 1][2][k - 1]);
    			Update(F[i][3][k], A[i][1] + A[i][2] + F[i - 1][3][k]);
    
    			if (k >= 1)
    				for (int j = 0; j <= 4; j++)
    					Update(F[i][4][k], A[i][1] + A[i][2] + F[i - 1][j][k - 1]);
    			Update(F[i][4][k], A[i][1] + A[i][2] + F[i - 1][4][k]);
    		}
    	int ans = MINF;
    	for (int i = 0; i <= 4; i++)
    		Update(ans, F[TotRow][i][K]);
    	return ans;
    }
    
    int main()
    {
    	scanf("%d%d%d", &TotRow, &TotCol, &K);
    	for (int i = 1; i <= TotRow; i++)
    		for (int j = 1; j <= TotCol; j++)
    			scanf("%d", &A[i][j]);
    	if (TotCol == 1)
    		printf("%d
    ", DP1());
    	else
    		printf("%d
    ", DP2());
    }
    

      

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    
    const int MAX_ROW = 110, MAX_COL = 5, MAX_K = 15, MINF = 0xcfcfcfcf;
    int F[MAX_ROW][MAX_ROW][MAX_K];
    int A[MAX_ROW][MAX_COL];
    int Sum[MAX_ROW][MAX_COL];
    int TotRow, TotCol, K;
    
    void InitSum()
    {
    	for (int row = 1; row <= TotRow; row++)
    		for (int col = 1; col <= TotCol; col++)
    			Sum[row][col] = Sum[row - 1][col] + Sum[row][col - 1] - Sum[row - 1][col - 1] + A[row][col];
    }
    
    int GetSum(int row1, int col1, int row2, int col2)
    {
    	return Sum[row2][col2] - Sum[row2][col1 - 1] - Sum[row1 - 1][col2] + Sum[row1 - 1][col1 - 1];
    }
    
    int DP1()
    {
    	memset(F, 0xcf, sizeof(F));
    	F[0][0][0] = 0;
    	for (int i = 1; i <= TotRow; i++)
    	{
    		for (int k = 0; k <= K; k++)
    		{
    			F[i][0][k] = F[i - 1][0][k];
    			if (k >= 1)
    				for (int j = 1; j <= i; j++)
    					F[i][0][k] = max(F[i][0][k], F[j - 1][0][k - 1] + GetSum(j, 1, i, 1));
    		}
    	}
    	return F[TotRow][0][K];
    }
    
    int DP2()
    {
    	memset(F, 0xcf, sizeof(F));
    	F[0][0][0] = 0;
    	for (int i = 0; i <= TotRow; i++)
    	{
    		for (int j = 0; j <= TotRow; j++)
    		{
    			for (int k = 0; k <= K; k++)
    			{
    				if (i >= 1)
    					F[i][j][k] = max(F[i][j][k], F[i - 1][j][k]);
    				if (j >= 1)
    					F[i][j][k] = max(F[i][j][k], F[i][j - 1][k]);
    				if (k >= 1)
    				{
    					for (int i1 = 1; i1 <= i; i1++)
    						F[i][j][k] = max(F[i][j][k], F[i1 - 1][j][k - 1] + GetSum(i1, 1, i, 1));
    					for (int j1 = 1; j1 <= j; j1++)
    						F[i][j][k] = max(F[i][j][k], F[i][j1 - 1][k - 1] + GetSum(j1, 2, j, 2));
    					if (i == j)
    						for (int i1 = 1; i1 <= i; i1++)
    							F[i][j][k] = max(F[i][j][k], F[i1 - 1][i1 - 1][k - 1] + GetSum(i1, 1, j, 2));
    				}
    			}
    		}
    	}
    	return F[TotRow][TotRow][K];
    }
    
    int main()
    {
    	scanf("%d%d%d", &TotRow, &TotCol, &K);
    	for (int i = 1; i <= TotRow; i++)
    		for (int j = 1; j <= TotCol; j++)
    			scanf("%d", &A[i][j]);
    	InitSum();
    	if (TotCol == 1)
    		printf("%d
    ", DP1());
    	else
    		printf("%d
    ", DP2());
    }
    

      

  • 相关阅读:
    java如何遍历Enumeration
    attachEvent与addEventlistener兼容性
    jquery如何把一个html元素替换成另外一个html元素?
    Struts2使用struts标签判断变量是否为空的写法
    ORACLE WITH AS 用法,创建临时表
    Hibernate传递list参数的例子
    java利用反射机制获取list中的某个字段并以list形式返回
    深入Golang之sync.Pool详解
    深入Golang调度器之GMP模型
    软技能,程序员编程之外的升值之道!【转】
  • 原文地址:https://www.cnblogs.com/headboy2002/p/9382092.html
Copyright © 2020-2023  润新知