USACO 之 Section 1.4 More Search Techniques （已解决）

Posted 2020-06-18 JmingS

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。

*/

 

  1 #define _CRT_SECURE_NO_WARNINGS
  2 #define HOME
  3 #define _CRTDBG_MAP_ALLOC
  4 #include <crtdbg.h>
  5 
  6 /*
  7 ID: Jming
  8 PROG: ariprog
  9 LANG: C++
 10 */
 11 #include <iostream>
 12 #include <fstream>
 13 #include <sstream>
 14 #include <cstdlib>
 15 #include <cstdio>
 16 #include <cstddef>
 17 #include <iterator>
 18 #include <algorithm>
 19 #include <string>
 20 #include <locale>
 21 #include <cmath>
 22 #include <vector>
 23 #include <cstring>
 24 #include <map>
 25 #include <utility>
 26 #include <queue>
 27 #include <stack>
 28 #include <set>
 29 #include <functional>
 30 using namespace std;
 31 typedef pair<int, int> PII;
 32 typedef long long int64;
 33 const int INF = 0x3f3f3f3f;
 34 const int modPrime = 3046721;
 35 const double eps = 1e-9;
 36 const int MaxA = 250 * 250 + 250 * 250 + 10;
 37 const int MaxB = 10010;
 38 
 39 int N, M;
 40 
 41 priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > pair_ans;
 42 bool isLegal[MaxA];
 43 
 44 int limVal;
 45 
 46 
 47 void ini()
 48 {
 49     /*
 50     预处理，找到all integers of the form p^2 + q^2 
 51     (where p and q are non-negative integers)
 52     */
 53     fill(isLegal, isLegal + MaxA + 1, false);
 54     int doubleVal[255];
 55     for (int i = 0; i <= M; ++i)
 56     {
 57         doubleVal[i] = i*i;
 58     }
 59     for (int i = 0; i <= M; ++i)
 60     {
 61         for (int j = 0; j <= M; ++j)
 62         {
 63             isLegal[doubleVal[i] + doubleVal[j]] = true;
 64         }
 65     }
 66 
 67 }
 68 
 69 
 70 void Solve()
 71 {
 72     for (int a = 0; a <= limVal; ++a)
 73     {
 74         for (int b = 1; b <= limVal; ++b)
 75         {
 76             if (a + (N - 1)*b > limVal)
 77             {
 78                 /*
 79                 判断以a为等差数列的第一个值，b为等差数列的公差时，
 80                 会不会超过(p^2 + q^2)的上界
 81                 */
 82                 break;
 83             }
 84             int nowVal = a;
 85             int cnt = 0;
 86             while (true)
 87             {
 88                 if (isLegal[nowVal])
 89                 {
 90                     ++cnt;
 91                 }
 92                 else
 93                 {
 94                     break;
 95                 }
 96                 if (cnt == N)
 97                 {
 98                     pair<int, int> pII;
 99                     pII.first = b;
100                     pII.second = a;
101                     pair_ans.push(pII);
102                     break;
103                 }
104                 nowVal += b;
105             }
106         }
107     }
108 }
109 
110 
111 void printAns()
112 {
113     if (pair_ans.empty())
114     {
115         puts("NONE");
116     }
117     else
118     {
119         while (!pair_ans.empty())
120         {
121             printf("%d %d\n", pair_ans.top().second, pair_ans.top().first);
122             pair_ans.pop();
123         }
124     }
125 }
126 
127 int main()
128 {
129 #ifdef HOME
130     freopen("in", "r", stdin);
131     //freopen("out", "w", stdout);
132 #endif
133 
134     freopen("ariprog.in", "r", stdin);
135     freopen("ariprog.out", "w", stdout);
136 
137     scanf("%d %d", &N, &M);
138     limVal = M*M + M*M;
139     ini();
140     Solve();
141     printAns();
142 
143 
144 #ifdef HOME
145     cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
146     _CrtDumpMemoryLeaks();
147 #endif
148     return 0;
149 }

 

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
*/

 

  1 #define _CRT_SECURE_NO_WARNINGS
  2 #define HOME
  3 #define _CRTDBG_MAP_ALLOC
  4 #include <crtdbg.h>
  5 /*
  6 ID: Jming
  7 PROG: milk3
  8 LANG: C++
  9 */
 10 #include <iostream>
 11 #include <fstream>
 12 #include <sstream>
 13 #include <cstdlib>
 14 #include <cstdio>
 15 #include <cstddef>
 16 #include <iterator>
 17 #include <algorithm>
 18 #include <string>
 19 #include <locale>
 20 #include <cmath>
 21 #include <vector>
 22 #include <cstring>
 23 #include <map>
 24 #include <utility>
 25 #include <queue>
 26 #include <stack>
 27 #include <set>
 28 #include <functional>
 29 using namespace std;
 30 typedef pair<int, int> PII;
 31 typedef long long int64;
 32 const int INF = 0x3f3f3f3f;
 33 const int modPrime = 3046721;
 34 const double eps = 1e-9;
 35 const int MaxN = 25;
 36 const int MaxM = 10010;
 37 
 38 int A, B, C;
 39 bool isVisited[25][25][25];
 40 set<int> ansSet;
 41 
 42 void Solve(int a, int b, int c)
 43 {
 44     if (!isVisited[a][b][c])
 45     {
 46         isVisited[a][b][c] = true;
 47         if (0 == a)
 48         {
 49             ansSet.insert(c);
 50         }
 51 
 52         // C->A
 53         if (c >= (A - a))
 54         {
 55             Solve(A, b, c - (A - a));
 56         }
 57         else
 58         {
 59             Solve(a + c, b, 0);
 60         }
 61 
 62         // C->B
 63         if (c >= (B - b))
 64         {
 65             Solve(a, B, c - (B - b));
 66         }
 67         else
 68         {
 69             Solve(a, b + c, 0);
 70         }
 71 
 72         // B->A
 73         if (b >= (A - a))
 74         {
 75             Solve(A, b - (A - a), c);
 76         }
 77         else
 78         {
 79             Solve(a + b, 0, c);
 80         }
 81 
 82         // B->C
 83         if (b >= (C - c))
 84         {
 85             Solve(a, b - (C - c), C);
 86         }
 87         else
 88         {
 89             Solve(a, 0, c + b);
 90         }
 91 
 92         // A->B
 93         if (a >= (B - b))
 94         {
 95             Solve(a - (B - b), B, c);
 96         }
 97         else
 98         {
 99             Solve(0, b + a, c);
100         }
101 
102         // A->C
103         if (a >= (C - c))
104         {
105             Solve(a - (C - c), b, C);
106         }
107         else
108         {
109             Solve(0, b, c + a);
110         }
111     }
112 }
113 
114 
115 int main()
116 {
117 #ifdef HOME
118     freopen("in", "r", stdin);
119     //freopen("out", "w", stdout);
120 #endif
121 
122     freopen("milk3.in", "r", stdin);
123     freopen("milk3.out", "w", stdout);
124 
125     scanf("%d %d %d", &A, &B, &C);
126     memset(isVisited, false, sizeof(isVisited));
127     Solve(0, 0, C);
128     for (set<int>::iterator it = ansSet.begin(); it != ansSet.end(); )
129     {
130         printf("%d", *it);
131         ++it;
132         if (it != ansSet.end())
133         {
134             printf(" ");
135         }
136     }
137     printf("\n");
138 
139 #ifdef HOME
140     cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
141     _CrtDumpMemoryLeaks();
142 #endif
143     return 0;
144 }