poj 2955 区间dp

题意：求最大括号匹配

链接：点我

区间dp水题，但也是很经典，区间dp都是由此延伸出来的，越是基础题越是需要勤加练习

Sample Input

((()))
()()()
([]])
)[)(
([][][)
end

Sample Output

6
6
4
0
6

 

2015-07-19：专题复习纪念

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<queue>
 7 #include<map>
 8 using namespace std;
 9 #define MOD 1000000007
10 const int INF=0x3f3f3f3f;
11 const double eps=1e-5;
12 typedef long long ll;
13 #define cl(a) memset(a,0,sizeof(a))
14 #define ts printf("*****\n");
15 const int MAXN=1005;
16 int n,m,tt;
17 char s[MAXN];
18 int dp[MAXN][MAXN];
19 int flag[MAXN][MAXN];
20 int main()
21 {
22     int i,j,k;
23     #ifndef ONLINE_JUDGE
24     freopen("1.in","r",stdin);
25     #endif
26     while(gets(s))
27     {
28         if(s[0]=='e')   break;
29         int len=strlen(s);
30         for(i=0;i<len;i++)  dp[i][i]=0;
31         for(int d=1;d<len;d++)
32         {
33             for(i=0;i+d<len;i++)
34             {
35                 j=i+d;
36                 dp[i][j]=-999999;
37                 if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']'))
38                 {
39                     dp[i][j]=dp[i+1][j-1]+1;
40                 }
41                 for(k=i;k<j;k++)
42                 {
43                     if(dp[i][j]<dp[i][k]+dp[k+1][j])
44                     {
45                         dp[i][j]=dp[i][k]+dp[k+1][j];
46                     }
47                 }
48             }
49         }
50         printf("%d\n",dp[0][len-1]*2);
51     }
52 }
View Code

 

 

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<queue>
 7 #include<map>
 8 using namespace std;
 9 #define MOD 1000000007
10 const int INF=0x3f3f3f3f;
11 const double eps=1e-5;
12 typedef long long ll;
13 #define cl(a) memset(a,0,sizeof(a))
14 #define ts printf("*****\n");
15 const int MAXN=1005;
16 int n,m,tt;
17 int dp[MAXN][MAXN];
18 int main()
19 {
20     int i,j,k;
21     #ifndef ONLINE_JUDGE
22     freopen("1.in","r",stdin);
23     #endif
24     char s[MAXN];
25     while(scanf("%s",s)!=EOF)
26     {
27         if(s[0]=='e')   break;
28         int len=strlen(s);
29         cl(dp);
30         for(int d=1;d<len;d++)
31         {
32             for(i=0;d+i<len;i++)
33             {
34                 j=d+i;
35                 if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']'))  dp[i][j]=dp[i+1][j-1]+2;
36                 for(k=i;k<=j;k++)
37                 {
38                     dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
39                 }
40             }
41         }
42         printf("%d\n",dp[0][len-1]);
43     }
44 }

 

posted @ 2015-05-06 17:23  miao_a_miao  阅读(181)  评论(0)  编辑  收藏  举报