poj 2241 叠方块 基本动态规划

题意：给定n种block，每种block有无限多个，每个block有x, y, z三个属性

要求叠起来，使得在满足下面的长和宽严格大于上面的，情况下，高度最高

解法：

d[i] 表示 以第i个物品为顶能达到的最大高度

转移方程 d[i] = max{ d[i] ， d[x] + height(x) 其中x要满足题目要求约束，遍取 0 - n - 1}

max(d[x])就是答案

#include <iostream>
#include <vector>
#include <map>
#include <list>
#include <set>
#include <deque>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <cstdio>
#include <iomanip>
#include <cmath>
#include <cstdio>
#include <iostream>
#include <string>
#include <sstream>
#include <cstring>
#include <queue>
using namespace std;

///宏定义
const int  INF = 990000000;
const int maxn = 500 ;
const int MAXN = maxn;
///全局变量 和 函数
//int T;
int max(int a, int b)
{
return a > b ? a : b;
}

int n;
struct BLOCK
{
int x, y, z;
bool operator < (const BLOCK& t) const
{
return (x * y) > (t.x * t.y);
}
};
BLOCK blocks[maxn * 5];
int d[maxn * 5];
bool vis[maxn * 5];
int cnt;
int dp(int k)
{
if (vis[k])
return d[k];
vis[k] = true;
int i, j;
int nowy, nowx, mh, maxheight;
//以x, y为底

mh = blocks[k].z;
nowy = blocks[k].y;
nowx = blocks[k].x;
maxheight = mh;
for (i = cnt - 1; i >= 0; i--)
{
if ( (nowy < blocks[i].y && nowx < blocks[i].x) || (nowx < blocks[i].y && nowy < blocks[i].x) ) //写错了好几次
{
maxheight = max(maxheight, dp(i) + mh);
}
}
return d[k] = maxheight;
}
int main()
{
///变量定义
int i, j;
int cases = 1;
while(1)
{
memset(vis, false, sizeof(vis));
scanf("%d", &n);
if (n == 0)
break;
cnt = 0;
for (i = 0; i < n; i++)
{
int x, y, z;
scanf("%d %d %d", &x, &y, &z);

blocks[cnt].x = x;
blocks[cnt].y = y;
blocks[cnt].z = z;
cnt++;

blocks[cnt].x = x;
blocks[cnt].y = z;
blocks[cnt].z = y;
cnt++;

blocks[cnt].x = y;
blocks[cnt].y = z;
blocks[cnt].z = x;
cnt++;
}
//		sort(blocks, blocks + cnt);
//		vis[0] = true;
//		d[0] = blocks[0].z;
int ans = -1;
for (i = cnt - 1; i >= 0; i--)
{
ans = max(ans, dp(i));
}
printf("Case %d: maximum height = %d\n", cases++, ans);
}

///结束
return 0;
}

在此基础上上更改一下细节处理，对于block的面积从大到小进行排序

那么d[i] = max{ d[i], d[k] + height[k] k < i }

速度会更快

#include <iostream>
#include <vector>
#include <map>
#include <list>
#include <set>
#include <deque>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <cstdio>
#include <iomanip>
#include <cmath>
#include <cstdio>
#include <iostream>
#include <string>
#include <sstream>
#include <cstring>
#include <queue>
using namespace std;

///宏定义
const int  INF = 990000000;
const int maxn = 500 ;
const int MAXN = maxn;
///全局变量 和 函数
//int T;
int max(int a, int b)
{
return a > b ? a : b;
}

int n;
struct BLOCK
{
int x, y, z;
bool operator < (const BLOCK& t) const
{
return (x * y) > (t.x * t.y);
}
};
BLOCK blocks[maxn * 5];
int d[maxn * 5];
bool vis[maxn * 5];
int cnt;
int dp(int k)
{
if (vis[k])
return d[k];
vis[k] = true;
int i, j;
int nowy, nowx, mh, maxheight;
//以x, y为底

mh = blocks[k].z;
nowy = blocks[k].y;
nowx = blocks[k].x;
maxheight = mh;
for (i = k - 1; i >= 0; i--)
{
if ( (nowy < blocks[i].y && nowx < blocks[i].x) || (nowx < blocks[i].y && nowy < blocks[i].x) ) //写错了好几次
{
maxheight = max(maxheight, dp(i) + mh);
}
}
return d[k] = maxheight;
}
int main()
{
///变量定义
int i, j;
int cases = 1;
while(1)
{
memset(vis, false, sizeof(vis));
scanf("%d", &n);
if (n == 0)
break;
cnt = 0;
for (i = 0; i < n; i++)
{
int x, y, z;
scanf("%d %d %d", &x, &y, &z);

blocks[cnt].x = x;
blocks[cnt].y = y;
blocks[cnt].z = z;
cnt++;

blocks[cnt].x = x;
blocks[cnt].y = z;
blocks[cnt].z = y;
cnt++;

blocks[cnt].x = y;
blocks[cnt].y = z;
blocks[cnt].z = x;
cnt++;
}
sort(blocks, blocks + cnt); //先进行排序
vis[0] = true;              //面积最大的作为顶的最大高度，记忆化搜索
d[0] = blocks[0].z;
int ans = -1;
for (i = cnt - 1; i >= 0; i--)
{
ans = max(ans, dp(i));
}
printf("Case %d: maximum height = %d\n", cases++, ans);
}

///结束
return 0;
}