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  • 河南省第八届ACM程序设计大赛

     

     

    A:挑战密室

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #include <stack>
    #include <map>
    #include <vector>
    #include <queue>
    using namespace std;
    typedef long long LL;
    #define N 600
    #define met(a, b) memset(a, b, sizeof(a))
    #define INF 0x3f3f3f3f
     
    char s[N];
     
    int Find(int x, int len)
    {
        int i;
     
        for(i=x; i<=len; i++)
            if(s[i]=='=')
            return i+1;
    }
    int Find2(int x, int len)
    {
        int i;
     
        for(i=x; i<=len; i++)
            if(s[i]=='+')
            return i-1;
        return len;
    }
     
     
    int Find1(int x)
    {
        int i, len = strlen(s)-1;
     
        for(i=x; i<len; i++)
            if(s[i]==')')
            return i-1;
     
        return len;
    }
     
    int Slove(int start, int e)
    {
        int k=0, sum=0, pre, i;
     
        for(i=start; i<=e; i++)
        {
            if(s[i]>='0' && s[i]<='9')
                k = k*10 + s[i]-'0';
            else break;
        }
     
        for( ;i<=e; i++)
        {
            if(s[i]=='N')
            {
                if(s[i+1]=='a')
                {
                    sum += 23;
                    pre = 23;
                    i++;
                }
                else
                {
                    sum += 14;
                    pre = 14;
                }
            }
            else if(s[i]=='C')
            {
                if(s[i+1]=='l')
                {
                    sum += 35;
                    pre = 35;
                    i++;
                }
                else if(s[i+1]=='a')
                {
                    sum += 40;
                    pre = 40;
                    i++;
                }
                else
                {
                    sum += 12;
                    pre = 12;
                }
            }
            else if(s[i]=='O')
            {
                sum += 16;
                pre = 16;
            }
            else if(s[i]=='S')
            {
                sum += 32;
                pre = 32;
            }
            else if(s[i]=='H')
            {
                sum += 2;
                pre = 2;
            }
            else if(s[i]=='A')
            {
                sum += 27;
                pre = 27;
                i++;
            }
            else if(s[i]=='Z')
            {
                sum += 65;
                pre = 65;
                i++;
            }
            else if(s[i]=='(')
            {
                int Index = Find1(i);
                int w = Slove(i+1, Index);
                sum += w;
                pre =  w;
                i = Index+1;
            }
            else
            {
                int z=0;
                while(s[i]>='0' && s[i]<='9')
                {
                    z = z*10 + s[i]-'0';
                    i++;
                }
                i--;
                sum += (z-1)*pre;
            }
        }
     
        if(k==0) return sum;
        else     return k*sum;
    }
     
    int main()
    {
        int T;
        scanf("%d", &T);
        while(T--)
        {
            int Index1, Index2, sum, len;
     
     
            scanf("%s", s);
     
            len = strlen(s)-1;
     
            Index1 = Find(0, len);
            Index2 = Find2(Index1, len);
            sum = Slove(Index1, Index2);
     
            printf("%04d
    ", sum);
        }
        return 0;
    }
    View Code

     

    B:最大岛屿

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #include <stack>
    #include <map>
    #include <vector>
    #include <queue>
    using namespace std;
    typedef long long LL;
    #define N 550
    #define met(a, b) memset(a, b, sizeof(a))
    #define INF 0x3f3f3f3f
     
    int s[N][N];
    int n, m, Max, dir[8][2]={{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
     
    int DFS(int x, int y)
    {
        int i, nx, ny, sum=0;
     
        s[x][y] = 0;
        for(i=0; i<8; i++)
        {
           nx = x + dir[i][0];
           ny = y + dir[i][1];
           if(nx>=0 && nx<n && ny>=0 && ny<m && s[nx][ny])
           {
              sum += DFS(nx, ny) + 1;
           }
        }
     
        return sum;
    }
     
    int main()
    {
        int T;
     
        while(scanf("%d%d%d", &n, &m, &T)!=EOF)
        {
            int i, j, ans, sum=0, Max=0;
     
            met(s, 0);
            for(i=0; i<n; i++)
            for(j=0; j<m; j++)
                scanf("%1d", &s[i][j]);
     
            for(i=0; i<n; i++)
            for(j=0; j<m; j++)
            {
                if(s[i][j]==1)
                {
                    ans = DFS(i, j) + 1;
                    ///printf("%d
    ", ans);
                    if(ans) sum++;
                    Max = max(Max, ans);
                }
            }
     
            printf("%d %d
    ", sum, Max*T);
        }
        return 0;
    }
    View Code

     

    C:最小换乘

    #include<iostream>
    #include<stdio.h>
    #include<algorithm>
    #include<math.h>
    #include<string.h>
    #include<string>
    #include<map>
    using namespace std;
    #define N 1305
    #define INF 0x3f3f3f3f
    #define met(a, b) memset(a, b, sizeof(a))
    
    int vis[N], G[N][N], n, dist[N];
    
    int Dij(int s)
    {
        met(vis, 0);
        dist[s] = 0;
        ///vis[s] = 1;
        for(int i=1; i<=n; i++)
        {
            int Min = INF, Index = -1;
            for(int j=1; j<=n; j++)
            {
                if(!vis[j] && Min>dist[j])
                {
                    Min = dist[j];
                    Index = j;
                }
            }
            if(Index == -1)break;
            vis[Index] = 1;
            for(int j=1; j<=n; j++)
            {
                if(!vis[j] && dist[j] > dist[Index]+G[Index][j])
                {
                    dist[j] = dist[Index] + G[Index][j];
                }
            }
        }
        return dist[n];
    }
    
    int main()
    {
        int m, T;
        scanf("%d", &T);
        while(T--)
        {
            scanf("%d %d", &m, &n);
    
            for(int i=1; i<=n; i++)
            {
                dist[i] = INF;
                for(int j=1; j<=n; j++)
                    G[i][j] = INF;
                G[i][i] = 0;
            }
    
            for(int i=1; i<=m; i++)
            {
                char ch; int k = 0, a[N];
    
                while(scanf("%d%c", &a[k], &ch),ch==' ')
                    k++;
    
                for(int j=0; j<=k; j++)
                {
                    for(int ii=j+1; ii<=k; ii++)
                        G[a[j]][a[ii]] = 1;
                }
            }
            int ans = Dij(1);
            if(ans == INF)
                puts("NO");
            else
                printf("%d
    ", ans-1);
        }
        return 0;
    }
    
    
    /**
    
    2
    3 7
    6 7
    4 7 3 6
    2 1 3 5
    2 6
    1 3 5
    2 6 4 3
    
    */
    View Code

    D:引水工程

    #include<iostream>
    #include<stdio.h>
    #include<algorithm>
    #include<math.h>
    #include<string.h>
    #include<string>
    #include<map>
    using namespace std;
    #define N 305
    #define INF 0x3f3f3f3f
    #define met(a, b) memset(a, b, sizeof(a))
     
    int vis[N], G[N][N], n, dist[N];
     
    int Prim(int s)
    {
        for(int i=1; i<=n; i++)
            dist[i] = G[s][i];
        met(vis, 0);
        vis[s] = 1;
        int ans = 0;
        for(int i=1; i<=n; i++)
        {
            int Min = INF, Index = -1;
            for(int j=1; j<=n; j++)
            {
                if(!vis[j] && dist[j]<Min)
                {
                    Min = dist[j];
                    Index = j;
                }
            }
            if(Index == -1)break;
            vis[Index] = 1;
            ans += Min;
            for(int j=1; j<=n; j++)
            {
                if(!vis[j] && dist[j]>G[Index][j])
                {
                    dist[j] = G[Index][j];
                }
            }
        }
        return ans;
    }
     
    int main()
    {
        int w, T;
        scanf("%d", &T);
        while(T--)
        {
            scanf("%d", &n);
     
            for(int i=0; i<=n; i++)
            {
                for(int j=0; j<=n; j++)
                {
                    G[i][j] = G[j][i] = INF;
                }
            }
     
            for(int i=1; i<=n; i++)
            {
                scanf("%d", &w);
                G[0][i] = G[i][0] = w;
            }
            for(int i=1; i<=n; i++)
            {
                for(int j=1; j<=n; j++)
                {
                    scanf("%d", &w);
                    G[i][j] = w;
                }
            }
            int ans = Prim(0);
            printf("%d
    ", ans);
        }
        return 0;
    }
    View Code

     

    F:Distribution

    #include<iostream>
    #include<stdio.h>
    #include<algorithm>
    #include<math.h>
    #include<string.h>
    #include<string>
    #include<map>
    using namespace std;
    #define N 10100
    #define met(a, b) memset(a, b, sizeof(a))
     
    int main()
    {
        int n, m, x[N], y[N], a, b;
        while(scanf("%d %d", &n, &m)!=EOF)
        {
            for(int i=0; i<n; i++)
            {
                scanf("%d %d", &x[i], &y[i]);
            }
            for(int i=1; i<=m; i++)
            {
                scanf("%d %d", &a, &b);
                int cnt1 = 0, cnt2 = 0;
                for(int j=0; j<n; j++)
                {
                    if((x[j]>a && y[j]>b) || (x[j]<a && y[j]<b))
                        cnt1++;
                    else if((x[j]<a && y[j]>b) || (x[j]>a && y[j]<b))
                        cnt2++;
                }
                printf("%d
    ", cnt1-cnt2);
            }
        }
        return 0;
    }
    View Code

     

    G:Interference Signal

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #include <stack>
    #include <map>
    #include <vector>
    #include <queue>
    using namespace std;
    typedef long long LL;
    #define N 1100
    #define met(a, b) memset(a, b, sizeof(a))
    #define INF 0x3f3f3f3f
    
    int a[N];
    
    int main()
    {
        int T;
        scanf("%d", &T);
        while(T--)
        {
            int n, k, i, j;
            double Max=0, ans;
    
            scanf("%d%d", &n, &k);
    
            met(a, 0);
            for(i=0; i<n; i++)
                scanf("%d", &a[i]);
    
            for(i=0; i<n; i++)
            {
                int sum = 0;
                for(j=i; j<n; j++)
                {
                    sum += a[j];
                    if((j-i+1)>=k)
                    {
                         ans = (sum*1.0)/(j-i+1);
                         Max = max(Max, ans);
                    }
                }
            }
    
            printf("%d
    ", (int)(Max*1000));
        }
        return 0;
    }
    View Code

     

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  • 原文地址:https://www.cnblogs.com/YY56/p/5523775.html
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