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Problem Description
Search is important in the acm algorithm. When you want to solve a problem by using the search method, try to cut is very important.
Now give you a number sequence, include n (<=100) integers, each integer not bigger than 2^31, you want to find the first P subsequences that is not decrease (if total subsequence W is smaller than P, than just give the first W subsequences). The order of subsequences is that: first order the length of the subsequence. Second order the subsequence by lexicographical. For example initial sequence 1 3 2 the total legal subsequences is 5. According to order is {1}; {2}; {3}; {1,2}; {1,3}. If you also can not understand , please see the sample carefully. |
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Input
The input contains multiple test cases.
Each test case include, first two integers n, P. (1<n<=100, 1<p<=100000). |
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Output
For each test case output the sequences according to the problem
description. And at the end of each case follow a empty line.
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Sample Input
3 5 1 3 2 3 6 1 3 2 4 100 1 2 3 2 |
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Sample Output
1 2 3 1 2 1 3 1 2 3 1 2 1 3 1 2 3 1 2 1 3 2 2 2 3 1 2 2 1 2 3 Hint
Hint : You must make sure each subsequence in the subsequences is unique. |
思路：正解应该是dfs，但是我为了偷懒，想开一个巨大的数组bfs，没想到hdu卡内存啊 啊啊啊！！害得我调试了好久好久！该死！
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <set> 7 using namespace std; 8 9 const int maxn=101; 10 const int limit=101000; 11 const int kkk=limit+100; 12 struct qq 13 { 14 int a,b; 15 bool friend operator < (const qq &c,const qq &d) 16 { 17 if (c.a==d.a) 18 return c.b<d.b; 19 return c.a<d.a; 20 } 21 }q[maxn]; 22 23 int n,p,tail,baktail,head,sum1,rec,bt,lh0,lh1,h,t; 24 short int l[kkk][2]; 25 short int x[kkk][maxn]; 26 int cnt=0,sum[maxn],s[maxn][maxn],pos[maxn][maxn]; 27 bool flag,flag2,flag3; 28 29 void close() 30 { 31 fclose(stdin); 32 fclose(stdout); 33 exit(0); 34 } 35 36 void print() 37 { 38 for (int i=h+1;i<=t;i++) 39 { 40 for (int j=1;j<l[i][1];j++) 41 { 42 printf("%d ",q[x[i][j]].a); 43 } 44 printf("%d\n",q[x[i][l[i][1]]].a); 45 } 46 } 47 48 bool judge() 49 { 50 p--; 51 if (p<=0) 52 { 53 return true; 54 } 55 return false; 56 } 57 58 59 60 void work() 61 { 62 head=0;tail=0;flag=false,flag2=true,flag3=false,cnt=0; 63 memset(l,0,sizeof(l)); 64 memset(x,0,sizeof(x)); 65 h=0; 66 t=0; 67 for (int i=1;i<=sum[0];i++) 68 { 69 t++; 70 if (judge()) 71 { 72 print(); 73 return; 74 } 75 x[i][1]=s[0][i]; 76 l[i][0]=pos[0][i]; 77 l[i][1]=1; 78 } 79 print(); 80 h=0; 81 p++; 82 while (h!=t) 83 { 84 bt=t; 85 while (h!=bt) 86 { 87 h=h % limit +1; 88 lh0=l[h][0]; 89 lh1=l[h][1]; 90 for (int i=1;i<=sum[lh0];i++) 91 { 92 if (judge()) 93 { 94 return; 95 } 96 t=t % limit +1; 97 memcpy(x[t],x[h],(lh1+1)*sizeof(int)); 98 l[t][1]=lh1+1; //多了一个数 99 l[t][0]=pos[lh0][i]; 100 x[t][l[t][1]]=s[lh0][i]; 101 for (int j=1;j<l[t][1];j++) 102 { 103 printf("%d ",q[x[t][j]].a); 104 } 105 printf("%d\n",q[x[t][l[t][1]]].a); 106 } 107 } 108 } 109 print(); 110 } 111 112 void init() 113 { 114 freopen("seq.in","r",stdin); 115 freopen("seq.out","w",stdout); 116 while (scanf("%d%d",&n,&p)!=EOF) 117 { 118 memset(q,0,sizeof(q)); 119 for (int i=1;i<=n;i++) 120 { 121 scanf("%d",&q[i].a); 122 q[i].b=i; 123 } 124 sort(q+1,q+n+1); 125 memset(s,0,sizeof(s)); 126 memset(pos,0,sizeof(pos)); 127 memset(sum,0,sizeof(sum)); 128 for (int i=0;i<=n;i++) 129 { 130 rec=-1000; 131 for (int j=i+1;j<=n;j++) 132 { 133 if (q[i].b<q[j].b && rec!=q[j].a) 134 { 135 flag=false; 136 if (flag) continue; 137 sum[i]++; 138 s[i][sum[i]]=j; 139 pos[i][sum[i]]=j; 140 rec=q[j].a; 141 } 142 } 143 } 144 work(); 145 printf("\n"); 146 } 147 } 148 149 int main () 150 { 151 init(); 152 close(); 153 return 0; 154 }