USACO 之 Section 1.4 More Search Techniques （已解决）

Arithmetic Progressions:

/*
需要注意两个可以优化的地方：
1）预处理，找到all integers of the form p^2 + q^2
　　(where p and q are non-negative integers)
2）判断以a为等差数列的第一个值，b为等差数列的公差时，会不会超过(p^2 + q^2)的上界

代码中相应位置做了说明。

*/

 

#define _CRT_SECURE_NO_WARNINGS
#define HOME
#define _CRTDBG_MAP_ALLOC
#include <crtdbg.h>

/*
ID: Jming
PROG: ariprog
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <sstream>
#include <cstdlib>
#include <cstdio>
#include <cstddef>
#include <iterator>
#include <algorithm>
#include <string>
#include <locale>
#include <cmath>
#include <vector>
#include <cstring>
#include <map>
#include <utility>
#include <queue>
#include <stack>
#include <set>
#include <functional>
using namespace std;
typedef pair<int, int> PII;
typedef long long int64;
const int INF = 0x3f3f3f3f;
const int modPrime = 3046721;
const double eps = 1e-9;
const int MaxA = 250 * 250 + 250 * 250 + 10;
const int MaxB = 10010;

int N, M;

priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > pair_ans;
bool isLegal[MaxA];

int limVal;


void ini()
{
    /*
    预处理，找到all integers of the form p^2 + q^2 
    (where p and q are non-negative integers)
    */
    fill(isLegal, isLegal + MaxA + 1, false);
    int doubleVal[255];
    for (int i = 0; i <= M; ++i)
    {
        doubleVal[i] = i*i;
    }
    for (int i = 0; i <= M; ++i)
    {
        for (int j = 0; j <= M; ++j)
        {
            isLegal[doubleVal[i] + doubleVal[j]] = true;
        }
    }

}


void Solve()
{
    for (int a = 0; a <= limVal; ++a)
    {
        for (int b = 1; b <= limVal; ++b)
        {
            if (a + (N - 1)*b > limVal)
            {
                /*
                判断以a为等差数列的第一个值，b为等差数列的公差时，
                会不会超过(p^2 + q^2)的上界
                */
                break;
            }
            int nowVal = a;
            int cnt = 0;
            while (true)
            {
                if (isLegal[nowVal])
                {
                    ++cnt;
                }
                else
                {
                    break;
                }
                if (cnt == N)
                {
                    pair<int, int> pII;
                    pII.first = b;
                    pII.second = a;
                    pair_ans.push(pII);
                    break;
                }
                nowVal += b;
            }
        }
    }
}


void printAns()
{
    if (pair_ans.empty())
    {
        puts("NONE");
    }
    else
    {
        while (!pair_ans.empty())
        {
            printf("%d %d\n", pair_ans.top().second, pair_ans.top().first);
            pair_ans.pop();
        }
    }
}

int main()
{
#ifdef HOME
    freopen("in", "r", stdin);
    //freopen("out", "w", stdout);
#endif

    freopen("ariprog.in", "r", stdin);
    freopen("ariprog.out", "w", stdout);

    scanf("%d %d", &N, &M);
    limVal = M*M + M*M;
    ini();
    Solve();
    printAns();


#ifdef HOME
    cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
    _CrtDumpMemoryLeaks();
#endif
    return 0;
}

 

Mother's Milk:

/*
题意关键：
　　FJ pours milk from one bucket to another until
　　the second bucket is filled or the first bucket is empty.
　　（每一次倒牛奶都遵守这个原则）

深搜，判重
*/

 

#define _CRT_SECURE_NO_WARNINGS
#define HOME
#define _CRTDBG_MAP_ALLOC
#include <crtdbg.h>
/*
ID: Jming
PROG: milk3
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <sstream>
#include <cstdlib>
#include <cstdio>
#include <cstddef>
#include <iterator>
#include <algorithm>
#include <string>
#include <locale>
#include <cmath>
#include <vector>
#include <cstring>
#include <map>
#include <utility>
#include <queue>
#include <stack>
#include <set>
#include <functional>
using namespace std;
typedef pair<int, int> PII;
typedef long long int64;
const int INF = 0x3f3f3f3f;
const int modPrime = 3046721;
const double eps = 1e-9;
const int MaxN = 25;
const int MaxM = 10010;

int A, B, C;
bool isVisited[25][25][25];
set<int> ansSet;

void Solve(int a, int b, int c)
{
    if (!isVisited[a][b][c])
    {
        isVisited[a][b][c] = true;
        if (0 == a)
        {
            ansSet.insert(c);
        }

        // C->A
        if (c >= (A - a))
        {
            Solve(A, b, c - (A - a));
        }
        else
        {
            Solve(a + c, b, 0);
        }

        // C->B
        if (c >= (B - b))
        {
            Solve(a, B, c - (B - b));
        }
        else
        {
            Solve(a, b + c, 0);
        }

        // B->A
        if (b >= (A - a))
        {
            Solve(A, b - (A - a), c);
        }
        else
        {
            Solve(a + b, 0, c);
        }

        // B->C
        if (b >= (C - c))
        {
            Solve(a, b - (C - c), C);
        }
        else
        {
            Solve(a, 0, c + b);
        }

        // A->B
        if (a >= (B - b))
        {
            Solve(a - (B - b), B, c);
        }
        else
        {
            Solve(0, b + a, c);
        }

        // A->C
        if (a >= (C - c))
        {
            Solve(a - (C - c), b, C);
        }
        else
        {
            Solve(0, b, c + a);
        }
    }
}


int main()
{
#ifdef HOME
    freopen("in", "r", stdin);
    //freopen("out", "w", stdout);
#endif

    freopen("milk3.in", "r", stdin);
    freopen("milk3.out", "w", stdout);

    scanf("%d %d %d", &A, &B, &C);
    memset(isVisited, false, sizeof(isVisited));
    Solve(0, 0, C);
    for (set<int>::iterator it = ansSet.begin(); it != ansSet.end(); )
    {
        printf("%d", *it);
        ++it;
        if (it != ansSet.end())
        {
            printf(" ");
        }
    }
    printf("\n");

#ifdef HOME
    cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
    _CrtDumpMemoryLeaks();
#endif
    return 0;
}