5.13解题报告

　　300分

　　T1写了差不多30~40分钟

　　T2写了不到5min (当时怀疑有坑..)

　　T3推了大概1个多小时的式子, 然后又加上调试差不多一个半小时

　　时间分配就这样..感觉T2出的太过简单了..直接是个模板

　　T1 并查集 + 乱搞

　　T2 快速幂模板

　　T3 Dp

T1 : codevs 2796 最小完全图

二次联通门 : codevs 2796 最小完全图

/*
    codevs 2796 最小完全图
    
    并查集 + 乱搞
    
    用并查集维护图
    给并查集带个集合中元素大小的权(即该并查集内共有多少点)
    
    最终的答案就是要合并的两个集合中点数的乘积-1
    因为两个集合中的点都不同, 所以可连得边数就是点数的乘积(size[x] * size[y])
    因为这两个集合中已经有当前边了, 所以注意要减去1
    
    由于是求最小
    那么边权就是当前边的边权加1啦
    (感觉自己说的很迷..)  
    注意每次合并时记得要按祖先来合并并查集的大小 
*/
#include <algorithm>
#include <cstdio>

#define Max 100090

void read (int &now)
{
    now = 0;
    register char word = getchar ();
    while (word < '0' || word > '9')
        word = getchar ();
    while (word >= '0' && word <= '9')
    {
        now = now * 10 + word - '0';
        word = getchar ();
    }
} 

int N;

long long Answer;

class Unio_Find_Set_Type
{
    private :
        
        int father[Max];
        int size[Max];
        
    public : 
        
        void clear (int N)
        {
            for (int i = 1; i <= N; i++)
            {
                father[i] = i;
                size[i] = 1; 
            }
        }
        
        int Find (int x)
        {
            return father[x] == x ? x : father[x] = Find (father[x]);
        }
        
        inline void Unio (int x, int y, int dis)
        {
            Answer += (long long) (dis + 1) * ((long long) size[x] * (long long) size[y] - 1);
            size[x] += size[y];
            father[y] = x;
        } 
};

Unio_Find_Set_Type ZlycerQan;

class Make_Graph_Type
{
    private :    
        
        struct Edges
        {
            int from;
            int to;
            int dis;
            
            bool operator < (const Edges a) const
            {
                return dis < a.dis;
            }
        }
    
        edge[Max];
    
        int Edge_Count;
        int now_1, now_2;
        
    public :
        
        inline void Add_Edge (int from, int to, int dis)
        {
            Answer += dis;
            Edge_Count++;
            edge[Edge_Count].from = from;
            edge[Edge_Count].to = to;
            edge[Edge_Count].dis = dis;    
        }
        
        void Make ()
        {
            std :: sort (edge + 1, edge + N);
            for (int i = 1; i < N; i++)
            {
                now_1 = ZlycerQan.Find (edge[i].from);
                now_2 = ZlycerQan.Find (edge[i].to);
                ZlycerQan.Unio (now_1, now_2, edge[i].dis); 
            }
        }
};

Make_Graph_Type Make;

int main (int argc, char *argv[])
{
    read (N);
    int x, y, z;
    for (int i = 1; i < N; i++)
    {
        read (x);
        read (y);
        read (z);
        Make.Add_Edge (x, y, z); 
    }
    ZlycerQan.clear (N); 
    Make.Make (); 
    printf ("%lld", Answer);
    return 0;
}

T2 : codevs 1497 取余运算

二次联通门 : codevs 1497 取余运算

/*
    codevs 1497 取余运算
    
    快速幂模板..
    不知道为何不能直接用printf 直接把整个式子输出...
    就只能比较麻烦, 分开输出了.. 
*/
#include <cstdio>

void read (long long &now)
{
    now = 0;
    register char word = getchar ();
    while (word < '0' || word > '9')
        word = getchar ();
    while (word >= '0' && word <= '9')
    {
        now = now * 10 + word - '0';
        word = getchar ();
    }
}

long long Fast_Pow (long long now, long long p, long long Mod)
{
    long long Answer = 1;
    while (p)
    {
        if (p & 1)
            Answer = (Answer * now) % Mod;
        now = (now * now) % Mod;
        p >>= 1;
    }
    return Answer;
}

int main (int argc, char *argv[])
{
    long long x, p, Mod;
    read (x);
    read (p);
    read (Mod);
    printf ("%lld^", x);
    printf ("%lld mod", p);
    printf (" %lld=", Mod);
    printf ("%lld", Fast_Pow (x, p, Mod));
    return 0;
}

T3 : codevs 1257 打砖块

二次联通门 : codevs 1257 打砖块

/*
    codevs 1257 打砖块
    
    DP
    
    状态数很好确定
    dp[i][j][k] 表示 打到第i行第j列是第k个被打掉的最大价值
    那么方程就是 
    dp[i][j][k] = max (dp[i][j][k], dp[i + 1][p][k - j] + map[j][i])
    map[i][j]是个前缀和
    用来记录(1 ~ i)行中第j列的价值总和
    p是枚举当前列中的位置
     
    若砖块不是按照顺序打掉的(即类似于一个三角形), 则新开一列
    第0列 即dp[i][0][k], 存的是不打第i列的砖块
    打第i-1列中砖块的最大值
    这样递推一下就可以..
    注意边界条件, 和每一行中的k值的正确求出 
     
*/
#include <cstdlib>
#include <cstdio>
#include <ctime>

#define Max 105
#define V N - i + 1

void read (int &now)
{
    now = 0;
    register char word = getchar ();
    while (word < '0' || word > '9')
        word = getchar ();
    while (word >= '0' && word <= '9')
    {
        now = now * 10 + word - '0';
        word = getchar ();
    }
}

inline int max (int a, int b)
{
    return a > b ? a : b;
}

int N, M;
int dp[Max][Max][Max * 10];
int map[Max][Max];

int main (int argc, char *argv[])
{
    read (N);
    read (M);
    for (int i = 1; i <= N; i++)
        for (int j = 1; j <= V; j++)
        {
            read (map[i][j]);
            map[i][j] += map[i - 1][j];
        }
    int Answer = 0, now;
    for (int i = N; i >= 1; i--)
        for (int j = 0; j <= V; j++)
        {
            now = 0;
            for (int q = 1; q <= j; q++)
                now += q;
            if (j == 0)
                now = 1;
            for (int k = now; k <= M; k++)
                for (int p = j - 1; p <= V; p++)
                    if (p >= 0)
                        dp[i][j][k] = max (dp[i][j][k], dp[i + 1][p][k - j] + map[j][i]);
                    else
                        dp[i][j][k] = max (dp[i][j][k], dp[i + 1][0][k - j] + map[j][i]);
        }
    for (int i = 1; i <= N; i++)
        for (int j = 0; j <= V; j++)
            Answer = max (Answer, dp[i][j][M]);
    printf ("%d", Answer);
    return 0;
}