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 }

 

posted @ 2013-04-22 21:57  發_  阅读(243)  评论(0)  编辑  收藏  举报