C++博客-止于自娱-随笔分类-动态规划

USACO 4.1 Beef McNuggets

YZY — Sun, 12 Jul 2009 06:58:00 GMT

这题有些难。虽然知道是动态规划题，但是不知道要开多大的数组，后来看analysis用一个256大小的数组循环使用，方法很巧妙。
先将box进行排序。
如果box里面的数的最大公约数不为1的话，那么所有组成的数，只可能是这个公约数的倍数，因此没有上限，输出为0.
用last记录最小的“不能组成的数”。这样当last之后有boxs[0]个连续数都可以组成的话，那么所有的数都可以组成。
last+1...last+box[0]可以组成的话，那么每个数都加一个box[0],那么新一轮的box[0]个数也可以组成，以此类推。

#include <iostream>
#include <fstream>

using namespace std;

ifstream fin("nuggets.in");
ofstream fout("nuggets.out");

#ifdef _DEBUG
#define out cout
#define in cin
#else
#define out fout
#define in fin
#endif

int box_num;
int boxs[10];

bool ok[256];

int gcd(int a,int b)
{
    if(a<b) swap(a,b);

    int tmp;

    while(b!=0){
        tmp = a;
        a = b;
        b = tmp%b;
    }

    return a;
}

void solve()
{

    in>>box_num;
    for(int i=0;i<box_num;++i)
        in>>boxs[i];

    sort(&boxs[0],&boxs[box_num]);

    int t = boxs[0];

    for(int i=1;i<box_num;++i){
        t = gcd(t,boxs[i]);
    }

    if(t!=1){
        out<<0<<endl;
        return;
    }

    memset(ok,0,sizeof(ok));

    int last = 0;
    ok[0] = true;
    int i=0;

    while(true){
        if(ok[i%256]){
            ok[i%256] = 0;
            if(i-last>=boxs[0]){
                out<<last<<endl;
                return;
            }
            for(int x=0;x<box_num;++x){
                ok[(i+boxs[x])%256] = true;
            }
        }else{
            last = i;
        }
        ++i;
    }
}

int main(int argc,char *argv[])
{
    solve();
    return 0;
}

Beef McNuggets

Hubert Chen

Farmer Brown's cows are up in arms, having heard that McDonalds is considering the introduction of a new product: Beef McNuggets. The cows are trying to find any possible way to put such a product in a negative light.

One strategy the cows are pursuing is that of `inferior packaging'. ``Look,'' say the cows, ``if you have Beef McNuggets in boxes of 3, 6, and 10, you can not satisfy a customer who wants 1, 2, 4, 5, 7, 8, 11, 14, or 17 McNuggets. Bad packaging: bad product.''

Help the cows. Given N (the number of packaging options, 1 <= N <= 10), and a set of N positive integers (1 <= i <= 256) that represent the number of nuggets in the various packages, output the largest number of nuggets that can not be purchased by buying nuggets in the given sizes. Print 0 if all possible purchases can be made or if there is no bound to the largest number.

The largest impossible number (if it exists) will be no larger than 2,000,000,000.

PROGRAM NAME: nuggets

INPUT FORMAT

Line 1:	N, the number of packaging options
Line 2..N+1:	The number of nuggets in one kind of box

SAMPLE INPUT (file nuggets.in)

OUTPUT FORMAT

The output file should contain a single line containing a single integer that represents the largest number of nuggets that can not be represented or 0 if all possible purchases can be made or if there is no bound to the largest number.

SAMPLE OUTPUT (file nuggets.out)

YZY 2009-07-12 14:58 发表评论

USACO 3.3 A Game

YZY — Thu, 09 Jul 2009 05:50:00 GMT

动态规划法。
用dp[player][start][end]表示player在[start..end]会取得的最大值。
如果player==0，那么player有主动权，它要么选start,要么选end.显然，它要选使得对手得分最少的那一个。
当它选start时，对手所能到的最大值为dp[1][start+1][end]。当它选end时，对手所选的最大值是dp[1][start][end-1].
。所以我们选dp[1][start+1][end]和dp[1][start][end-1]中小的那一个。
如果player==1，那只能被动的等0先选了。1在剩下区段中，又作为先选的角色，即player0。
当只有一个数字的时候，player0就只有选这个，player1就没得选，返回0.
代码如下：

#include <iostream>
#include <fstream>

using namespace std;

ifstream fin("game1.in");
ofstream fout("game1.out");

#ifdef _DEBUG
#define out cout
#define in cin
#else
#define out fout
#define in fin
#endif

int dp[2][100][100];
int sequence[100];

int score(int player,int start,int end)
{
    if(dp[player][start][end]!=-1)
        return dp[player][start][end];

    if(start==end){
        if(player==0)
            dp[player][start][end] = sequence[start];
        else
            dp[player][start][end] = 0;
    }else{
        int t1 = score(0,start+1,end);
        int t2 = score(0,start,end-1);
        if(player==0){
            if(t1>t2){
                dp[player][start][end] = sequence[end]+score(1,start,end-1);
            }else{
                dp[player][start][end] = sequence[start]+score(1,start+1,end);
            }
        }else{
            if(t1>t2){
                dp[player][start][end] = score(0,start,end-1);
            }else{
                dp[player][start][end] = score(0,start+1,end);
            }
        }
    }

    return dp[player][start][end];
}

void solve()
{
    memset(dp,-1,sizeof(dp));

    int size;
    in>>size;

    for(int i=0;i<size;++i)
        in>>sequence[i];

    out<<score(0,0,size-1)<<" "<<score(1,0,size-1)<<endl;
}

int main(int argc,char *argv[])
{
    solve();
    return 0;
}

YZY 2009-07-09 13:50 发表评论

USACO 3.3 Home on the Range

YZY — Wed, 08 Jul 2009 09:44:00 GMT

dp题。用dp[i][j][m]表示左上角为(i,j)，边长为m的正方形是否完整。那么
如果dp[i][j][m]完整，则当且仅当dp[i][j][m-1],dp[i+1][j][m-1],dp[i][j+1][m-1],dp[i+1][j+1][m-1]的正方形也是完整的(画个图就很清晰了)。由于我们从上到下，从左到右扫描每个点，在每一轮i,j用过一次，就不会再使用，所以只需用二维数组保存dp[i][j]，即可。
时间复杂度为O(n^3),空间复杂度为O(n^2)。analysis中有个时间复杂度为O(n^2),空间O(n)的解法。

#include <iostream>
#include <fstream>

using namespace std;

ifstream fin("range.in");
ofstream fout("range.out");

#ifdef _DEBUG
#define out cout
#define in cin
#else
#define out fout
#define in fin
#endif

bool dp[250][250];
int side_len;

void input()
{
    in>>side_len;

    char t;

    for(int i=0;i<side_len;++i){
        for(int j=0;j<side_len;++j){
            while( in.get(t)&&isspace(t) );
            dp[i][j] = t-'0';
        }
    }
}

void solve()
{
    input();

    for(int w = 2;w<=side_len;++w){
        int cnt = 0;

        for(int i=0;i<side_len;++i){
            for(int j=0;j<side_len;++j){
                if(i+w<=side_len&&j+w<=side_len){
                    dp[i][j] = dp[i][j]&&dp[i+1][j]&&dp[i][j+1]&&dp[i+1][j+1];
                    if(dp[i][j])
                        cnt++;
                }
            }
        }

        if(cnt!=0){
            out<<w<<" "<<cnt<<endl;
        }
    }
}

int main(int argc,char *argv[])
{
    solve();
    return 0;
}

YZY 2009-07-08 17:44 发表评论

USACO 3.3 Shopping Offers

YZY — Tue, 07 Jul 2009 11:16:00 GMT

背包问题，dp解决。
dp[i][j][k][m][n]表示第0-4种物品分别购买i-n个时，所需的最小费用。
那么对于每一个offer
dp[i][j][k][m][n] = min ( dp[i][j][k][m][n], dp[i-o[0]][j-o[1]][k-o[2]][m-o[3]][n-o[4]]+offer.cost);

analysis中提出用最短路径的方法来解，思路很巧妙。把”装有不同种数和数量的物品“的篮子看作结点，把offer作为边(把购买一件物品的原始价格看成一种退化的offer)，把价格作为边长度。这样就转换成从篮子(0,0,0,0,0)到所求结点的最短路径问题。

代码如下:

#include <iostream>
#include <fstream>

using namespace std;

ifstream fin("shopping.in");
ofstream fout("shopping.out");

#ifdef _DEBUG
#define out cout
#define in cin
#else
#define out fout
#define in fin
#endif

int dp[6][6][6][6][6];

struct offer{
    int pro_num;
    int price;
    int product[5];
    int amount[5];
};

int offer_num;
offer offers[100];
int map[1000]; //product id到”按物品出现顺序所给的编号“的映射
int price[1000];//product id对应的物品价格
int product_num;//物品总数目
int products[5];//存放product id
int amount[5];//product 所需的数量

void solve()
{
    in>>offer_num;

    for(int i=0;i<offer_num;++i){
        in>>offers[i].pro_num;
        for(int j=0;j<offers[i].pro_num;++j){
            in>>offers[i].product[j]>>offers[i].amount[j];
        }
        in>>offers[i].price;
    }

    int pro_idx = 0;

    in>>product_num;

    for(int i=0;i<product_num;++i){
        in>>products[i];
        in>>amount[i];
        in>>price[i];
        map[ products[i] ] = i;
    }

    //没有折扣时的价格
    for(int i=0;i<=5;++i)
        for(int j=0;j<=5;++j)
            for(int k=0;k<=5;++k)
                for(int m=0;m<=5;++m)
                    for(int n=0;n<=5;++n){
                        dp[i][j][k][m][n] =
                            price[0]*i+
                            price[1]*j+
                            price[2]*k+
                            price[3]*m+
                            price[4]*n;
                    }

    for(int i=0;i<=5;++i)
        for(int j=0;j<=5;++j)
            for(int k=0;k<=5;++k)
                for(int m=0;m<=5;++m)
                    for(int n=0;n<=5;++n){
                        int tmp[5];
                        for(int s=0;s<offer_num;++s){

                            memset(tmp,0,sizeof(tmp));
                            for(int t=0;t<offers[s].pro_num;++t){
                                tmp[map[offers[s].product[t]]] = offers[s].amount[t];
                            }

                            if(i>=tmp[0]&&j>=tmp[1]&&k>=tmp[2]&&m>=tmp[3]&&n>=tmp[4]){
                                dp[i][j][k][m][n] = min( dp[i][j][k][m][n],
                                        dp[i-tmp[0]][j-tmp[1]][k-tmp[2]][m-tmp[3]][n-tmp[4]]+
                                        offers[s].price);
                            }
                        }
                    }

    out<<dp[amount[0]][amount[1]][amount[2]][amount[3]][amount[4]]<<endl;
}

int main(int argc,char *argv[])
{
    solve();
    return 0;
}

上面代码有个严重的bug，谢谢网友“我也凑热闹"指出。由于map所有值都为0。所以未在商品列表中出现的商品的map值都为0，即都映射为第一个商品。现改成将map初始化为-1，并增加判断语句。此外将初始化dp的语句合并到后面，以简化代码。

#include <iostream>
#include <fstream>

using namespace std;

ifstream fin("shopping.in");
ofstream fout("shopping.out");

#ifdef _DEBUG
#define out cout
#define in cin
#else
#define out fout
#define in fin
#endif

int dp[6][6][6][6][6];

struct offer{
    int pro_num;
    int price;
    int product[5];
    int amount[5];
};

int offer_num;
offer offers[100];
int map[1000];
int price[1000];
int product_num;
int products[5];
int amount[5];

void solve()
{
    in>>offer_num;

    for(int i=0;i<offer_num;++i){
        in>>offers[i].pro_num;
        for(int j=0;j<offers[i].pro_num;++j){
            in>>offers[i].product[j]>>offers[i].amount[j];
        }
        in>>offers[i].price;
    }

    int pro_idx = 0;

    in>>product_num;

    //2009.07.27修改
    memset(map,-1,sizeof(map));

    for(int i=0;i<product_num;++i){
        in>>products[i];
        in>>amount[i];
        in>>price[i];
        map[ products[i] ] = i;
    }

    for(int i=0;i<=5;++i)
        for(int j=0;j<=5;++j)
            for(int k=0;k<=5;++k)
                for(int m=0;m<=5;++m)
                    for(int n=0;n<=5;++n){

                         dp[i][j][k][m][n] =
                            price[0]*i+
                            price[1]*j+
                            price[2]*k+
                            price[3]*m+
                            price[4]*n;

                        int tmp[5];
                        for(int s=0;s<offer_num;++s){
                            memset(tmp,0,sizeof(tmp));
                            for(int t=0;t<offers[s].pro_num;++t){
                                //2009.07.27修改
                                if( map[offers[s].product[t]]!=-1)
                                    tmp[map[offers[s].product[t]]] = offers[s].amount[t];
                            }

                            if(i>=tmp[0]&&j>=tmp[1]&&k>=tmp[2]&&m>=tmp[3]&&n>=tmp[4]){
                                dp[i][j][k][m][n] = min( dp[i][j][k][m][n],
                                        dp[i-tmp[0]][j-tmp[1]][k-tmp[2]][m-tmp[3]][n-tmp[4]]+
                                        offers[s].price);
                            }
                        }
                    }

    out<<dp[amount[0]][amount[1]][amount[2]][amount[3]][amount[4]]<<endl;
}

int main(int argc,char *argv[])
{
    solve();
    return 0;
}

YZY 2009-07-07 19:16 发表评论

USACO 3.1 Stamps

YZY — Tue, 30 Jun 2009 12:37:00 GMT

摘要: 动态规划题阅读全文

YZY 2009-06-30 20:37 发表评论

USACO 3.1 Score Inflation

YZY — Mon, 29 Jun 2009 14:21:00 GMT

摘要: 背包问题阅读全文

YZY 2009-06-29 22:21 发表评论

USACO 2.3 Money Systems

YZY — Thu, 25 Jun 2009 14:16:00 GMT

摘要: 动态规划经典题阅读全文

YZY 2009-06-25 22:16 发表评论

USACO 2.3 Cow Pedigrees

YZY — Wed, 24 Jun 2009 12:55:00 GMT

摘要: memoization dp, 树的计数阅读全文

YZY 2009-06-24 20:55 发表评论

USACO 2.3 Longest Prefix

YZY — Tue, 23 Jun 2009 13:13:00 GMT

摘要: Trie树+简单DP 阅读全文

YZY 2009-06-23 21:13 发表评论