TC SRM 584 DIV2

250pt:

水题set处理。

500pt：

题意：

给你一个图，每条边关联的两点为朋友，题目要求假设x的金钱为y，则他的左右的朋友当中的钱数z，取值为y - d <= z <= y + d.求使得任意两点的最大金钱差值，若果是inf输出-1.

思路：
求任意两点的最短的的最大值即可，比赛时不知道哪地方写搓了，直接被系统样例给虐了，老师这么悲剧500有思路能写，老师不仔细哎..

floyd求任意两点的最短距离好写一些。

#include <iostream>

#include <cstdio>

#include <cmath>

#include <vector>

#include <cstring>

#include <algorithm>

#include <string>

#include <set>

#include <functional>

#include <numeric>

#include <sstream>

#include <stack>

#include <map>

#include <queue>



#define CL(arr, val)    memset(arr, val, sizeof(arr))



#define lc l,m,rt<<1

#define rc m + 1,r,rt<<1|1

#define pi acos(-1.0)

#define ll __int64

#define L(x)    (x) << 1

#define R(x)    (x) << 1 | 1

#define MID(l, r)   (l + r) >> 1

#define Min(x, y)   (x) < (y) ? (x) : (y)

#define Max(x, y)   (x) < (y) ? (y) : (x)

#define E(x)        (1 << (x))

#define iabs(x)     (x) < 0 ? -(x) : (x)

#define OUT(x)  printf("%I64d\n", x)

#define lowbit(x)   (x)&(-x)

#define keyTree (chd[chd[root][1]][0])

#define Read()  freopen("din.txt", "r", stdin)

#define Write() freopen("dout.txt", "w", stdout);





#define M 25000

#define N 207



using namespace std;





const int inf = 0x7f7f7f7f;

const int mod = 1000000007;



int f[N][N];

int n;

void floyd()

{

    for (int k = 1; k <= n; ++k)

    {

        for (int i = 1; i <= n; ++i)

        {

            for (int j = 1; j <= n; ++j)

            {

                if (f[i][k] != inf && f[k][j] != inf && f[i][j] > f[i][k] + f[k][j])

                {

                    f[i][j] = f[i][k] + f[k][j];

                }

            }

        }

    }



}

class Egalitarianism

{

    public:

    int maxDifference(vector <string> isF, int d)

    {

        n = isF.size();

        for (int i = 1; i <= n; ++i)

        {

            for (int j = 1; j <= n; ++j) f[i][j] = inf;

        }

        for (int i = 0; i < n; ++i)

        {

            for (int j = 0; j < n; ++j)

            {

                if (isF[i][j] == 'Y')

                {

                    f[i + 1][j + 1] = 1;

                }

            }

        }

        floyd();

        int Ma = 0;

        for (int i = 1; i <= n; ++i)

        {

            for (int j = 1; j <= n; ++j)

            {

                if (i == j) continue;

                Ma = max(Ma,f[i][j]);

            }

        }

        if (Ma == inf) return -1;

        else return Ma*d;

    }

};

View Code

 

100pt：

题意：

有n个城市，给出每个城市的类型kind[i]表示i城市属于kind[i]类，然后给出已经发现的类型found[i]表示发现了found[i]类型， 然后给出k求满足有k个城市，并且这k个城市包含了m中已将发现的类型。（k个城市都是从已经发现的类型里面选的） 也即：知道m种数的每种数个数，然后将这些数放入K个箱子里面连，每个箱子只能放一个，要求放完后这k个箱子的数的种数为m

思路：
才开始想用组和公式方看看怎么放，可是那样考虑会有很多重复，而且不好去重。

dp其实很简单，可是就是想不到，弱逼的DP啊。  dp[i][j] 表示一共放了j个箱子，并且放了i种 则dp[i][j] += dp[i - 1][j - p]*c[found[i]][p];

#include <iostream>

#include <cstdio>

#include <cmath>

#include <vector>

#include <cstring>

#include <algorithm>

#include <string>

#include <set>

#include <functional>

#include <numeric>

#include <sstream>

#include <stack>

#include <map>

#include <queue>



#define CL(arr, val)    memset(arr, val, sizeof(arr))



#define lc l,m,rt<<1

#define rc m + 1,r,rt<<1|1

#define pi acos(-1.0)

#define ll long long

#define L(x)    (x) << 1

#define R(x)    (x) << 1 | 1

#define MID(l, r)   (l + r) >> 1

#define Min(x, y)   (x) < (y) ? (x) : (y)

#define Max(x, y)   (x) < (y) ? (y) : (x)

#define E(x)        (1 << (x))

#define iabs(x)     (x) < 0 ? -(x) : (x)

#define OUT(x)  printf("%I64d\n", x)

#define lowbit(x)   (x)&(-x)

#define keyTree (chd[chd[root][1]][0])

#define Read()  freopen("din.txt", "r", stdin)

#define Write() freopen("dout.txt", "w", stdout);





#define M 25000

#define N 107



using namespace std;





const int inf = 0x7f7f7f7f;

const int mod = 1000000007;

ll dp[N][N];

ll c[55][55];

void init()

{

    for (int i = 0; i <= 50; ++i)

    {

        c[i][i] = c[i][0] = 1;

    }

    for (int i = 2; i <= 50; ++i)

    {

        for (int j = 1; j < i; ++j)

        {

            c[i][j] = c[i - 1][j] + c[i - 1][j - 1];

        }

    }

}

class Excavations2

{

    public:

    int num[N];



    long long count(vector <int> kind, vector <int> found, int K)

    {

        init();

        int n = kind.size();

        int m = found.size();

        CL(num,0);



        for (int i = 0; i < n; ++i) num[kind[i]]++;

        CL(dp,0); dp[0][0] = 1;



        for (int i = 0; i < m; ++i)

        {

            for (int j = 0; j <= K; ++j)

            {

                for (int p = 1; p <= num[found[i]]; ++p)

                {

                    if (j - p >= 0)

                    dp[i + 1][j] += dp[i][j - p]*c[num[found[i]]][p];

                }

            }

        }

        return dp[m][K];

    }

};

View Code