24 | 9月 | 2015 | 超客的水墨烟雨

poj2385

maksyuki 发表于 oj 分类，标签: DP-线性DP

24 9月 2015

Apple Catching

It is a little known fact that cows love apples. Farmer John has two apple trees (which are conveniently numbered 1 and 2) in his field, each full of apples. Bessie cannot reach the apples when they are on the tree, so she must wait for them to fall. However, she must catch them in the air since the apples bruise when they hit the ground (and no one wants to eat bruised apples). Bessie is a quick eater, so an apple she does catch is eaten in just a few seconds.

Each minute, one of the two apple trees drops an apple. Bessie, having much practice, can catch an apple if she is standing under a tree from which one falls. While Bessie can walk between the two trees quickly (in much less than a minute), she can stand under only one tree at any time. Moreover, cows do not get a lot of exercise, so she is not willing to walk back and forth between the trees endlessly (and thus misses some apples).

Apples fall (one each minute) for T (1 <= T <= 1,000) minutes. Bessie is willing to walk back and forth at most W (1 <= W <= 30) times. Given which tree will drop an apple each minute, determine the maximum number of apples which Bessie can catch. Bessie starts at tree 1.

Input

* Line 1: Two space separated integers: T and W

* Lines 2..T+1: 1 or 2: the tree that will drop an apple each minute.

Output

* Line 1: The maximum number of apples Bessie can catch without walking more than W times.

Sample Input

7 2

Sample Output

Hint

INPUT DETAILS:

Seven apples fall - one from tree 2, then two in a row from tree 1, then two in a row from tree 2, then two in a row from tree 1. Bessie is willing to walk from one tree to the other twice.

OUTPUT DETAILS:

Bessie can catch six apples by staying under tree 1 until the first two have dropped, then moving to tree 2 for the next two, then returning back to tree 1 for the final two.

Source

USACO 2004 November

题目类型：线性DP

算法分析：dp[i][j]表示前i分钟恰好走j次所能够获得的最多苹果数，初始化dp数组全为0，状态转移方程为：

若第i分钟树1落下一个苹果

(1)此时j是奇数，表示当前人在树2下面，则dp[i][j] = max (dp[i-1][j-1], dp[i-1][j])

(2)此时j是偶数，表示当前人在树1下面，则dp[i][j] = 1 + max (dp[i-1][j-1], dp[i-1][j])

若第i分钟树2落下一个苹果

(1)此时j是奇数，表示当前人在树2下面，则dp[i][j] = 1 + max (dp[i-1][j-1], dp[i-1][j])

(2)此时j是偶数，表示当前人在树1下面，则dp[i][j] = max (dp[i-1][j-1], dp[i-1][j])

注意要特殊处理j = 0的情况！！！最后输出所有可能走的步数时的最大苹果数即可

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)
#define CPPFF ifstream cin ("aaa.txt")
#define LL long long
#define ULL unsigned long long
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl

const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 1e6 + 6;

int dp[1000+66][30+6], aa[maxn];

int main()
{
//    CFF;
    int t, w;
    while (scanf ("%d%d", &t, &w) != EOF)
    {
        for (int i = 1; i <= t; i++)
            scanf ("%d", &aa[i]);

        memset (dp, 0, sizeof (dp));
        for (int i = 1; i <= t; i++)
            for (int j = 0; j <= w; j++)
            {
                if (aa[i] == 1)
                {
                    if (j == 0)
                        dp[i][j] = dp[i-1][j] + 1;

                    else if (j & 1)
                        dp[i][j] = max (dp[i-1][j], dp[i-1][j-1]);

                    else
                        dp[i][j] = 1 + max (dp[i-1][j], dp[i-1][j-1]);
                }

                else
                {
                    if (j == 0)
                        dp[i][j] = dp[i-1][j];

                    else if (j & 1)
                        dp[i][j] = 1 + max (dp[i-1][j], dp[i-1][j-1]);

                    else
                        dp[i][j] = max (dp[i-1][j], dp[i-1][j-1]);
                }
            }


        printf ("%d\n", *max_element (dp[t], dp[t] + w + 1));
    }
    return 0;
}

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)

#define CPPFF ifstream cin ("aaa.txt")

#define LL long long

#define ULL unsigned long long

#define DB(ccc) cout << #ccc << " = " << ccc << endl

const int INF = 0x7FFFFFFF;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 1e6 + 6;

int dp[1000+66][30+6], aa[maxn];

int main()

{

// CFF;

int t, w;

while (scanf ("%d%d", &t, &w) != EOF)

{

for (int i = 1; i <= t; i++)

scanf ("%d", &aa[i]);

memset (dp, 0, sizeof (dp));

for (int i = 1; i <= t; i++)

for (int j = 0; j <= w; j++)

{

if (aa[i] == 1)

{

if (j == 0)

dp[i][j] = dp[i-1][j] + 1;

else if (j & 1)

dp[i][j] = max (dp[i-1][j], dp[i-1][j-1]);

else

dp[i][j] = 1 + max (dp[i-1][j], dp[i-1][j-1]);

}

else

{

if (j == 0)

dp[i][j] = dp[i-1][j];

else if (j & 1)

dp[i][j] = 1 + max (dp[i-1][j], dp[i-1][j-1]);

else

dp[i][j] = max (dp[i-1][j], dp[i-1][j-1]);

}

printf ("%d\n", *max_element (dp[t], dp[t] + w + 1));

}

return 0;

}

poj3659

maksyuki 发表于 oj 分类，标签: DP-树形DP, 图论-最小点支配

24 9月 2015

Cell Phone Network

Farmer John has decided to give each of his cows a cell phone in hopes to encourage their social interaction. This, however, requires him to set up cell phone towers on his N (1 ≤ N ≤ 10,000) pastures (conveniently numbered 1..N) so they can all communicate.

Exactly N-1 pairs of pastures are adjacent, and for any two pastures A and B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B) there is a sequence of adjacent pastures such that A is the first pasture in the sequence and B is the last. Farmer John can only place cell phone towers in the pastures, and each tower has enough range to provide service to the pasture it is on and all pastures adjacent to the pasture with the cell tower.

Help him determine the minimum number of towers he must install to provide cell phone service to each pasture.

Input

* Line 1: A single integer: N
* Lines 2..N: Each line specifies a pair of adjacent pastures with two space-separated integers: A and B

Output

* Line 1: A single integer indicating the minimum number of towers to install

Sample Input

1 3

5 2

4 3

3 5

Sample Output

Source

USACO 2008 January Gold

题目类型：最小点支配(树形DP)

算法分析：dp[u][1]表示以u为根的子树在u这个节点上面安排士兵所具有的最小点支配数，dp[u][2]表示以u为根的子树在u这个节点上不安排士兵，而被其子节点所支配的最小点支配数，dp[u][3]表示以u为根的子树在u这个节点上不安排士兵，而被其父节点所支配的最小点支配数。初始条件是dp[u][1] = 1, dp[u][2] = dp[u][3] = 0，状态转移方程为：
dp[u][1] += min (dp[v][1], dp[v][2], dp[v][3])

dp[u][3] += min (dp[v][1], dp[v][2])

(1)dp[u][2] = INF(若u是叶子节点)

(2)dp[u][2] = 若选择了的u所有子节点中有dp[v][1]则计算完后直接返回即可，反之则使用一个变量来记录dp[v][1] - dp[v][2]的最小值，最后循环完之后再加上这个变量的值

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)
#define CPPFF ifstream cin ("aaa.txt")
#define LL long long
#define ULL unsigned long long
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl

const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 1e6 + 6;

//分别表示顶点v。w和下一个邻接的位置
struct Node
{
    int v, w, next;
};

Node edge[maxn];
int head[maxn], edgelen;//分别表示顶点的表头节点和边的数量

//将所有的数组均初始化为-1
void Init ()
{
    memset (head, -1, sizeof (head));
    memset (edge, -1, sizeof (edge));
    edgelen = 0;
}

//使用头插法构建
void AddEdge (int u, int v, int w)
{
    edge[edgelen].v = v;
    edge[edgelen].w = w;
    edge[edgelen].next = head[u];
    head[u] = edgelen++;
}

bool vis[maxn];
int dp[maxn][6];

void dfs (int rt)
{
    vis[rt] = true;
    bool is_valid = true;
    int tt = INF;
    for (int i = head[rt]; i != -1; i = edge[i].next)
    {
        if (!vis[edge[i].v])
        {
            dfs (edge[i].v);
            dp[rt][1] += min (dp[edge[i].v][1], min (dp[edge[i].v][2], dp[edge[i].v][3]));
            dp[rt][3] += min (dp[edge[i].v][1], dp[edge[i].v][2]);
            if (dp[edge[i].v][1] <= dp[edge[i].v][2])
            {
                dp[rt][2] += dp[edge[i].v][1];
                is_valid = false;
            }

            else
            {
                dp[rt][2] += dp[edge[i].v][2];
                tt = min (tt, dp[edge[i].v][1] - dp[edge[i].v][2]);
            }
        }
    }

    if (is_valid)
        dp[rt][2] += tt;
}

int main()
{
//    CFF;
    int n;
    while (scanf ("%d", &n) != EOF)
    {
        Init ();
        for (int i = 1; i <= n; i++)
        {
            int u, v;
            scanf ("%d%d", &u, &v);
            AddEdge (u, v, 1);
            AddEdge (v, u, 1);
        }

        memset (vis, false, sizeof (vis));
        for (int i = 1; i <= n; i++)
            dp[i][2] = dp[i][3] = 0, dp[i][1] = 1;

        dfs (1);
        printf ("%d\n", min (dp[1][1], dp[1][2]));
    }
    return 0;
}

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#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)

#define CPPFF ifstream cin ("aaa.txt")

#define LL long long

#define ULL unsigned long long

#define DB(ccc) cout << #ccc << " = " << ccc << endl

const int INF = 0x7FFFFFFF;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 1e6 + 6;

//分别表示顶点v。w和下一个邻接的位置

struct Node

{

int v, w, next;

};

Node edge[maxn];

int head[maxn], edgelen;//分别表示顶点的表头节点和边的数量

//将所有的数组均初始化为-1

void Init ()

{

memset (head, -1, sizeof (head));

memset (edge, -1, sizeof (edge));

edgelen = 0;

}

//使用头插法构建

void AddEdge (int u, int v, int w)

{

edge[edgelen].v = v;

edge[edgelen].w = w;

edge[edgelen].next = head[u];

head[u] = edgelen++;

}

bool vis[maxn];

int dp[maxn][6];

void dfs (int rt)

{

vis[rt] = true;

bool is_valid = true;

int tt = INF;

for (int i = head[rt]; i != -1; i = edge[i].next)

{

if (!vis[edge[i].v])

{

dfs (edge[i].v);

dp[rt][1] += min (dp[edge[i].v][1], min (dp[edge[i].v][2], dp[edge[i].v][3]));

dp[rt][3] += min (dp[edge[i].v][1], dp[edge[i].v][2]);

if (dp[edge[i].v][1] <= dp[edge[i].v][2])

{

dp[rt][2] += dp[edge[i].v][1];

is_valid = false;

}

else

{

dp[rt][2] += dp[edge[i].v][2];

tt = min (tt, dp[edge[i].v][1] - dp[edge[i].v][2]);

}

if (is_valid)

dp[rt][2] += tt;

}

int main()

{

// CFF;

int n;

while (scanf ("%d", &n) != EOF)

{

Init ();

for (int i = 1; i <= n; i++)

{

int u, v;

scanf ("%d%d", &u, &v);

AddEdge (u, v, 1);

AddEdge (v, u, 1);

}

memset (vis, false, sizeof (vis));

for (int i = 1; i <= n; i++)

dp[i][2] = dp[i][3] = 0, dp[i][1] = 1;

dfs (1);

printf ("%d\n", min (dp[1][1], dp[1][2]));

}

return 0;

}

超客的水墨烟雨

poj2385

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