TC SRM 584 DIV2 | 易学教程

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

来源：https://www.cnblogs.com/E-star/p/3183417.html

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