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USACO 3.3 A Game

YZY — Thu, 09 Jul 2009 05:50:00 GMT

动态规划法�?br />用dp[player][start][end]表示player在[start..end]�?x��)取得的最大倹{�?br />如果player==0�Q�那么player有主动权�Q�它要么选start,要么选end.昄��Q�它要选��得对手得分最��的那一个�?br />当它选start�Ӟ��Ҏ(gu��)��所能到的最大��gؓ(f��)dp[1][start+1][end]。当它选end�Ӟ��Ҏ(gu��)��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�?br />如果player==1�Q�那只能被动的等0先选了�?在剩下区�D�中�Q�又作�ؓ(f��)先选的角色�Q�即player0�?br />当只有一个数字的时候，player0��只有选这个，player1��没得选，�q�回0.
代码如下�Q?br />

#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�ؓ(f��)(i,j)�Q�边长�ؓ(f��)m的正方�Ş是否完整。那�?br />如果dp[i][j][m]完整�Q�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�Ş也是完整�?��M��囑ְ�很清��C��)。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�ơ，��׃��?x��)再使用�Q�所以只需用二�l�数�l�保存dp[i][j]�Q�即可�?br />旉��复杂度�ؓ(f��)O(n^3),�I�间复杂度�ؓ(f��)O(n^2)。analysis中有个时间复杂度为O(n^2),�I�间O(n)的解法�?br />

#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�Q�dp解决�?br />dp[i][j][k][m][n]表示�W?-4�U�物品分别购买i-n个时�Q�所需的最��费用�?br />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�\径的�Ҏ(gu��)��来解�Q�思�\很��y妙。把”装有不同种数和数量的物品“的��子看作�l�点�Q�把offer作�ؓ(f��)�?把购��C��件物品的原始��h��看成一�U�退化的offer)�Q�把��h��作�ؓ(f��)辚w��度。这样就转换成从��子(0,0,0,0,0)到所求结点的最短�\径问题�?br />
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��序所�l�的�~�号“的映射
int price[1000];//product id对应的物品�h(hu��n)�?br />int product_num;//物品��L��?br />int products[5];//存放product id
int amount[5];//product 所需的数�?br />
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��h��
    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�Q�谢谢网友“我也凑热闹"指出。由于map所有值都�?。所以未在商品列表中出现的商品的map值都�?�Q�即都映��ؓ(f��)�W�一个商品。现�Ҏ(gu��)��map初始化�ؓ(f��)-1�Q��ƈ增加判断语句。此外将初始化dp的语句合�q�到后面�Q�以��化代码�?br />

#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: 动态规划经兔R�� 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, �?w��i)的计�?nbsp; 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�?��单DP 阅读全文

YZY 2009-06-23 21:13 发表评论