滑雪
[sol]
很经典的记忆化搜索的题目,记忆化的模板可以像树形dp的dfs那般实现,但是也
有不同,即:
dfs()
{
if (叶子节点)
{
dp初始化
return;
}
for (int i = 1; i < 4; i++)
{
if(判断)
{
dfs(下一节点)
状态转移
}
}
// 一般把四个方向的最优解处理给f[x0][y0]
}
[code]
#include<cstdio>
#include<iostream>
using namespace std;
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
const int N = 100 + 5;
int n, m;
int f[N][N], a[N][N];
bool isok(int x, int y)
{
return x >= 0 && x < n && y >= 0 && y < m;
}
void dfs(int x0, int y0)
{
f[x0][y0]=1;
int tot = 0;
for(int i = 0; i < 4; i++)
{
int x = x0 + dx[i], y = y0 + dy[i];
if (isok(x, y) && a[x0][y0] > a[x][y])
{
if (f[x][y] == 0)
dfs (x, y);
tot = max (tot, f[x][y]);
}
}
f[x0][y0] += tot;
//printf ("f[%d,%d]=%d\n", x0, y0, f[x0][y0]);
}
int main()
{
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;
for (int i =0; i < n; i++)
for (int j = 0; j < m; j++)
if (f[i][j] == 0)
{
dfs (i, j);
ans = max (ans, f[i][j]);
}
printf("%d", ans);
scanf("%d", &n);
return 0;
}
传纸条
[sol]
此题为noip真题,在座标系中向一个方向移动,用dp来搞是可以的,而一道道对角
线就是一个又一个阶段,第k步所能到达的点连线形成了的第k条对角线,每个点的
纵座标可以由对角线编号和横座标表示,这样令f[p,x1,y1] (x1< x2)代表第一个点在
(x1,p-x1),第二个点在(x2,p-x2)的最优解,顺着四个方向去转移方程就行,其实dp的第
一维可以用滚动数组去优化掉
#include<cstdio>
#include<iostream>
using namespace std;
const int N = 100 + 5;
int f[N * 2][N][N], a[N][N];
int n,m;
bool cal(int x, int y)
{
return x >=0 && x < n && y >= 0 && y < m;
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
scanf("%d", &a[i][j]);
f[1][0][1]=a[0][1] + a[1][0] + a[0][0];
for (int p = 1; p < n + m - 3; p++)
for (int x1 = 0; x1 <= min(p, n); x1++)
for (int x2 = x1 + 1; x2 <= min(p,n); x2++)
{
// printf("%d-----f[%d,%d]=%d\n", p, x1, x2, f[p][x1][x2]);
if (x1 != x2 && cal(x1 + 1, p - x1) && cal(x2, p - x2 + 1))
f[p + 1][x1 + 1][x2] = max (f[p + 1][x1 + 1][x2], f[p][x1][x2] + a[x1 + 1][p - x1] + a[x2][p - x2 + 1]);
if (cal(x1, p - x1 + 1) && cal(x2 + 1, p - x2))
f[p + 1][x1][x2 + 1] = max (f[p + 1][x1][x2 + 1], f[p][x1][x2] + a[x1][p - x1 + 1] + a[x2 + 1][p - x2]);
if (cal(x1 + 1, p - x1) && cal(x2 + 1, p - x2))
f[p + 1][x1 + 1][x2 + 1] = max (f[p + 1][x1 + 1][x2 + 1], f[p][x1][x2] + a[x1 + 1][p - x1] + a[x2 + 1][p - x2]);
if (cal(x1, p - x1 + 1) && cal(x2, p - x2 + 1))
f[p + 1][x1][x2] = max (f[p + 1][x1][x2], f[p][x1][x2] + a[x1][p - x1 + 1] + a[x2][p - x2 + 1]);
}
printf("%d",f[n + m - 3][n - 2][n - 1]);
return 0;
}
奶牛的锻炼
[sol]
直接按照题意去转移就可以,f[i,j]代表第i分钟疲劳值为j的最长距离,注意疲劳值为
0的情况,即f[i,0] = f[i-1,0]
#include<cstdio>
#include<iostream>
using namespace std;
const int N = 2000 + 5;
const int M = 500 + 5;
int f[N][M];
int d[N];
int main()
{
int n, m;
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)
scanf("%d", &d[i]);
for (int i = 1; i <= n; i++)
{
f[i][0]=f[i - 1][0];
for (int j = 1; j <= min(i - 1, m); j++)
f[i][0] = max (f[i][0], f[i - j][j]);
for (int j = 1; j <= min(i, m); j++)
f[i][j] = f[i - 1][j - 1] + d[i];
}
printf("%d",f[n][0]);
return 0;
}