Topcoder SRM 146

Posted 2020-08-06

Div1 300 RectangularGrid

题意：给定一个长方形，问包含有多少不是正方形的小长方形

题解：枚举小长方形的长宽即可

#line 2 "RectangularGrid.cpp"
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;

class RectangularGrid
{
    public:
    long long countRectangles(int width, int height)
    {
        LL ans=0;
        for (int i=1;i<=width;++i)
        {
            for (int j=1;j<=height;++j)
            {
                if (i==j) continue;
                int w=width-i+1;
                int h=height-j+1;
                ans+=w*h;
            }
        }
        return ans;
    }
};

#ifdef ex
int main()
{
    #ifdef ex1
    freopen ("in.txt","r",stdin);
    #endif

}
#endif

 

Div1 600 Masterbrain

题意：略

题解：主要是卡题意，以及“No digit in either a guess or a secret combination may be involved in giving more than one peg”这个条件的处理

#line 2 "Masterbrain.cpp"
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;

int ans;
int A[5];
int cnt[15];
int ok[5];
vector<vector<int> > G;
vector<int> v;
vector<int> B;
vector<int> W;

bool check()
{
    int wa=0;

    for (int i=0;i<G.size();++i)
    {
        memset(ok,0,sizeof(ok));

        vector<int>& v=G[i];
        int b=0;
        int w=0;
        for (int j=0;j<v.size();++j)
        {
            if (A[j]==v[j])
            {
                ++b;
                ok[j]=1;
            }
        }
        for (int j=0;j<v.size();++j)
        {
            if (A[j]!=v[j] && cnt[v[j]]!=0)
            {
                for (int jj=0;jj<v.size();++jj)
                {
                    if (A[jj]==v[j] && ok[jj]==0)
                    {
                        ok[jj]=1;
                        ++w;
                        break;
                    }
                }
            }
        }

        if (b!=B[i] || w!=W[i]) ++wa;
    }

    if (wa!=1) return false;
    else return true;
}

void dfs(int k)
{
    if (k==4)
    {
        if (check()) ++ans;
        return;
    }

    for (int i=1;i<=7;++i)
    {
        A[k]=i;
        ++cnt[i];
        dfs(k+1);
        --cnt[i];
    }
    return ;
}

class Masterbrain
{
    public:

    int possibleSecrets(vector <string> guesses, vector <string> results)
    {
        for (int i=0;i<guesses.size();++i)
        {
            string& tmp=guesses[i];
            v.clear();
            for (int j=0;j<tmp.length();++j)
            {
                int t=tmp[j]-‘0‘;
                v.push_back(t);
            }
            G.push_back(v);
        }

        for (int i=0;i<results.size();++i)
        {
            string& tmp=results[i];
            B.push_back(tmp[0]-‘0‘);
            W.push_back(tmp[3]-‘0‘);
        }

        ans=0;
        memset(cnt,0,sizeof(cnt));

        dfs(0);

        return ans;
    }
};

#ifdef ex
int main()
{
    #ifdef ex1
    freopen ("in.txt","r",stdin);
    #endif

}
#endif

 

Div2 1000 BridgeCrossing

题意：略

题解：状压DP

#line 2 "BridgeCrossing.cpp"
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;

const int maxn=1<<8;

int dp[maxn+5];
int sz;
vector<int> T;

int DP(int x)
{
    if (dp[x]!=-1) return dp[x];

    vector<int> v1,v2;
    for (int i=0;i<=sz-1;++i)
    {
        if (x>>i & 1) v1.push_back(i);
        else v2.push_back(i);
    }

    if (v1.size()==0) return dp[x]=0;
    else if (v1.size()==1) return dp[x]=T[v1[0]];
    else if (v1.size()==2) return dp[x]=max(T[v1[0]],T[v1[1]]);

    int q=0;
    int val=1e8;
    for (int i=0;i<v2.size();++i)
    {
        int& e=v2[i];
        if (T[e]<val)
        {
            val=T[e];
            q=e;
        }
    }

    dp[x]=1e8;
    for (int i=0;i<v1.size();++i)
    {
        for (int j=i+1;j<v1.size();++j)
        {
            int& e1=v1[i];
            int& e2=v1[j];

            int S=x&~(1<<e1);
            S=S&~(1<<e2);

            int t=T[e1]<T[e2]?e1:e2;
            int p=val<T[t]?q:t;

            S=S|(1<<p);
            dp[x]=min(dp[x],DP(S)+max(T[e1],T[e2])+T[p]);

            //printf("%d %d %d\n",x,S,q);
            //printf("%d %d %d %d\n",x,v1.size(),i,j);
        }
    }

    return dp[x];
}

class BridgeCrossing
{
    public:
    int minTime(vector <int> times)
    {
        sz=times.size();
        for (int i=0;i<=sz-1;++i) T.push_back(times[i]);

        memset(dp,-1,sizeof(dp));
        int ans=DP((1<<sz)-1);


        /*
        for (int i=0;i<=((1<<sz)-1);++i)
        {
            printf("%d %d\n",i,DP(i));
        }
        */
        return ans;
    }
};

#ifdef ex
int main()
{
    #ifdef ex1
    freopen ("in.txt","r",stdin);
    #endif

}
#endif