真题解析 Workpiece Match

Posted 2020-10-23 proscientist

零件匹配题目：

 

 

 

现在,有若干个零件，需要你匹配

求所有能匹配的组合中，最大的组合长度T_max (T释义见图4)，以及能匹配出??_max的组合个数n

例如:

有26个零件，分别是A-Z，其中能匹配的组合只有三组(A,B), (C,D), (E,F)

(A,B)匹配后,组合长度T=5

(C,D)匹配后,组合长度T=3

(E,F)匹配后,组合长度T=5

则 ??max=5, ??=2

 

输入

第一行一个整数t, 表示有t组测试数据,  1≤??≤50

每组第一行一个整数m,表示零件个数 , 0<??≤100,000

下面有m行,每行17个整数,表示零件数据

其中第一个表示零件本体长度L, 0<??≤100,000,000

剩余16个表示零件头矩阵,按从左到右,从上到下顺序排列，如下图。

对矩阵中每个元素??, 有 0≤??≤2

 

输出

最大的组合长度??T_max ，能匹配出??_max的组合个数n。

用空格分开，回车结束

 

待完善

 

参考代码：

  1 #include <stdio.h>
  2 #include <windows.h>
  3 #define MAXPAT 43046721
  4 #define MAXNUM 100005
  5 int T, N, ans, index = MAXNUM - 1, maxlen;
  6 int exp[16] = { 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907 };
  7 int key[MAXNUM];//优化
  8 struct Node{//存放最小的pat
  9     int len, num;
 10     int data[16], top;
 11     Node *next;
 12 }NIL, node[MAXNUM], *map[MAXPAT];
 13 int v[4][16] = {
 14     { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 },
 15     { 3, 7, 11, 15, 2, 6, 10, 14, 1, 5, 9, 13, 0, 4, 8, 12 },
 16     { 12, 8, 4, 0, 13, 9, 5, 1, 14, 10, 6, 2, 15, 11, 7, 3 },
 17     { 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0 }
 18 };
 19 int findVal(int a[16]){
 20     int ret = MAXPAT + 1, temp;
 21     for (int i = 0; i < 4; ++i){
 22         temp = 0;
 23         for (int j = 0; j < 16; ++j)
 24             temp = temp + a[j] * exp[v[i][j]];
 25         if (temp < ret)ret = temp;
 26     }
 27     return ret;
 28 }
 29 void getMaxLen(){
 30     maxlen = ans = 0;    
 31     int a[16];
 32     Node *lover;
 33     for (int i = 0; i <= index; ++i){
 34         int pos = key[i];
 35         if (map[pos]->len < 0)continue;
 36         for (int j = 0; j < 4; ++j)
 37         for (int k = 0; k < 4; ++k)
 38             a[j * 4 + k] = map[pos]->top - map[pos]->data[j * 4 + 3 - k];
 39         int pat = findVal(a);        
 40         if (map[pat]->len < 0){
 41             map[pos] = &NIL;
 42             continue;
 43         }
 44         if (pat < pos)continue;
 45         if(pat == pos){        
 46             if(map[pos]->num > 1){
 47                 if(map[pos]->len * 2 + map[pos]->top > maxlen){
 48                     maxlen = map[pos]->len * 2 + map[pos]->top;
 49                     ans = map[pos]->num / 2;
 50                 }
 51                 else if(map[pos]->len * 2 + map[pos]->top == maxlen)
 52                     ans += map[pos]->num / 2;    
 53                 map[pos] = &NIL;
 54                 continue;
 55             }
 56             else lover = map[pos]->next;
 57         }
 58         else{
 59             lover = map[pat];
 60             map[pat] = &NIL;
 61         }
 62         if(map[pos]->len + lover->len + map[pos]->top > maxlen){
 63             maxlen = map[pos]->len + lover->len + map[pos]->top;
 64             ans = map[pos]->num < lover->num ? map[pos]->num : lover->num;
 65         }
 66         else if(map[pos]->len + lover->len + map[pos]->top == maxlen)
 67             ans += map[pos]->num < lover->num ? map[pos]->num : lover->num;
 68         map[pos] = &NIL;
 69     }
 70 }
 71 int main(void){
 72     freopen("input.txt", "r", stdin);
 73     freopen("result.txt", "w", stdout);
 74     scanf("%d", &T);
 75     NIL.len = -1;
 76     NIL.next = &NIL;
 77     for (int i = 0; i < MAXPAT; ++i)map[i] = &NIL;
 78     DWORD start_time = GetTickCount();
 79     for (int tc = 1; tc <= T; ++tc){    
 80         index = -1;
 81         scanf("%d", &N);
 82         for (int i = 0; i < N; ++i){
 83             ++index;
 84             scanf("%d", &node[index].len);
 85             node[index].top = 0;
 86             node[index].num = 1;
 87             for (int j = 0; j < 16; ++j){
 88                 scanf("%d", &node[index].data[j]);
 89                 if (node[index].data[j] > node[index].top)
 90                     node[index].top = node[index].data[j];
 91             }
 92             int pat = findVal(node[index].data);
 93             key[index] = pat;
 94             if (map[pat]->len < node[index].len){//应该插入到头部
 95                 node[index].next = map[pat];
 96                 map[pat] = &node[index];
 97                 continue;
 98             }
 99             Node *cur = map[pat];
100             while (true){
101                 if (cur->len == node[index].len){
102                     ++cur->num;
103                     --index;
104                     break;
105                 }
106                 if (cur->next->len < node[index].len){
107                     node[index].next = cur->next;
108                     cur->next = &node[index];
109                     break;
110                 }
111                 cur = cur->next;
112             }
113         }
114         getMaxLen();
115         printf("#%d %d %d\\n", tc, maxlen, ans);
116     }
117     DWORD end_time = GetTickCount();
118     printf("%d ms\\n", end_time - start_time);
119     return 0;
120 }