hdu 2610 && hdu 2611(bfs)

hdu 2610题意：从数组中找出不超过P个序列，每个序列为不下降序列，输出按1）序列长度；2）每个数出现的先后

做法：bfs，题目要求完全符合bfs的性质。判重很关键，我用了比较水的set来判重。（用set<string>来判重wa，改用set<vector<int> >ac而且比前者快。）

343MS

 1 /*

 2  *Author:       Zhaofa Fang

 3  *Created time: 2013-04-22-11.50

 4  *Language:     C++

 5  */

 6 #include <cstdio>

 7 #include <cstdlib>

 8 #include <sstream>

 9 #include <iostream>

10 #include <cmath>

11 #include <cstring>

12 #include <algorithm>

13 #include <string>

14 #include <utility>

15 #include <vector>

16 #include <queue>

17 #include <map>

18 #include <set>

19 using namespace std;

20 

21 typedef long long ll;

22 #define DEBUG(x) cout<< #x << ':' << x << endl

23 #define FOR(i,s,t) for(int i = (s);i <= (t);i++)

24 #define FORD(i,s,t) for(int i = (s);i >= (t);i--)

25 #define REP(i,n) for(int i=0;i<n;i++)

26 #define REPD(i,n) for(int i=n-1;i>=0;i--)

27 #define PII pair<int,int>

28 #define PB push_back

29 #define MP make_pair

30 #define ft first

31 #define sd second

32 #define lowbit(x) (x&(-x))

33 #define INF (1<<30)

34 

35 

36 

37 int n,p;

38 int a[1011];

39 set<vector<int> >S;

40 

41 struct node{

42     vector<int> ss;

43     int tail,end;

44 };

45 

46 void bfs(){

47     S.clear();

48     queue<node>Q;

49     node now;

50     int cnt = 0;

51     FOR(i,1,n){

52         now.ss.clear();

53         now.ss.PB(a[i]);

54         if(S.count(now.ss))continue;

55         S.insert(now.ss);

56         now.end = i;

57         now.tail = a[i];

58         Q.push(now);

59         cnt ++;

60         cout<<a[i]<<endl;

61         if(cnt>=p)return;

62     }

63     while(!Q.empty()){

64         now = Q.front();

65         Q.pop();

66         FOR(i,now.end+1,n){

67             if(now.tail > a[i])continue;

68             node next;

69             next.ss = now.ss;

70             next.ss.PB(a[i]);

71             if(S.count(next.ss))continue;

72             next.end = i;

73             next.tail = a[i];

74             S.insert(next.ss);

75             int sz = next.ss.size();

76             REP(i,sz)printf("%d%c",next.ss[i],(i!=sz-1)?' ':'\n');

77             cnt++;

78             if(cnt>=p)return;

79             Q.push(next);

80         }

81     }

82 }

83 

84 int main(){

85     //freopen("in","r",stdin);

86     //freopen("out","w",stdout);

87     while(~scanf("%d%d",&n,&p)){

88         FOR(i,1,n)scanf("%d",a+i);

89         bfs();

90         puts("");

91     }

92     return 0;

93 }

 

hdu 2611 题意与上题的差异是输出规则2）按字典序输出

做法：我同样是用bfs和同样的判重方法。每个节点增加一个变量记录下标，先按数列的值的值排序，相同则按下标排（因为缺少了后者wa了几次）。在用bfs扩展的时候，要同时符合下标值和数值不小于前一个节点才扩展。

625MS

  1 /*

  2  *Author:       Zhaofa Fang

  3  *Created time: 2013-04-22-11.50

  4  *Language:     C++

  5  */

  6 #include <cstdio>

  7 #include <cstdlib>

  8 #include <sstream>

  9 #include <iostream>

 10 #include <cmath>

 11 #include <cstring>

 12 #include <algorithm>

 13 #include <string>

 14 #include <utility>

 15 #include <vector>

 16 #include <queue>

 17 #include <map>

 18 #include <set>

 19 using namespace std;

 20 

 21 typedef long long ll;

 22 #define DEBUG(x) cout<< #x << ':' << x << endl

 23 #define FOR(i,s,t) for(int i = (s);i <= (t);i++)

 24 #define FORD(i,s,t) for(int i = (s);i >= (t);i--)

 25 #define REP(i,n) for(int i=0;i<n;i++)

 26 #define REPD(i,n) for(int i=n-1;i>=0;i--)

 27 #define PII pair<int,int>

 28 #define PB push_back

 29 #define MP make_pair

 30 #define ft first

 31 #define sd second

 32 #define lowbit(x) (x&(-x))

 33 #define INF (1<<30)

 34 

 35 

 36 

 37 int n,p;

 38 set<vector<int> >S;

 39 

 40 struct seq{

 41     int x,pos;

 42 }a[111];

 43 bool cmp(seq x,seq y){

 44     if(x.x != y.x)return x.x<y.x;

 45     return x.pos<y.pos;

 46 }

 47 struct node{

 48     vector<int> ss;

 49     int tail,end,endpos;

 50 };

 51 

 52 void bfs(){

 53     S.clear();

 54     queue<node>Q;

 55     node now;

 56     int cnt = 0;

 57     FOR(i,1,n){

 58         now.ss.clear();

 59         now.ss.PB(a[i].x);

 60         if(S.count(now.ss))continue;

 61         S.insert(now.ss);

 62         now.end = i;

 63         now.tail = a[i].x;

 64         now.endpos = a[i].pos;

 65         Q.push(now);

 66         cnt ++;

 67         printf("%d\n",a[i]);

 68         if(cnt>=p)return;

 69     }

 70     while(!Q.empty()){

 71         now = Q.front();

 72         Q.pop();

 73         FOR(i,now.end+1,n){

 74             if(now.tail > a[i].x)continue;

 75             if(now.endpos > a[i].pos)continue;

 76             node next;

 77             next.ss = now.ss;

 78             next.ss.PB(a[i].x);

 79             if(S.count(next.ss))continue;

 80             next.end = i;

 81             next.tail = a[i].x;

 82             next.endpos = a[i].pos;

 83             S.insert(next.ss);

 84             int sz = next.ss.size();

 85             REP(i,sz)printf("%d%c",next.ss[i],(i!=sz-1)?' ':'\n');

 86             cnt++;

 87             if(cnt>=p)return;

 88             Q.push(next);

 89         }

 90     }

 91 }

 92 

 93 

 94 int main(){

 95     //freopen("in","r",stdin);

 96     //freopen("out","w",stdout);

 97     while(~scanf("%d%d",&n,&p)){

 98         FOR(i,1,n){

 99             scanf("%d",&a[i].x);

100             a[i].pos = i;

101         }

102         sort(a+1,a+n+1,cmp);

103         bfs();

104         puts("");

105     }

106     return 0;

107 }