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BNUOJ 1260 Brackets Sequence

2024-07-17 20:38:04   226人阅读

Brackets Sequence

Time Limit: 1000ms
Memory Limit: 65536KB
This problem will be judged on PKU. Original ID: 1141
64-bit integer IO format: %lld      Java class name: Main
Special Judge
 
Let us define a regular brackets sequence in the following way: 

1. Empty sequence is a regular sequence. 
2. If S is a regular sequence, then (S) and [S] are both regular sequences. 
3. If A and B are regular sequences, then AB is a regular sequence. 

For example, all of the following sequences of characters are regular brackets sequences: 

(), [], (()), ([]), ()[], ()[()] 

And all of the following character sequences are not: 

(, [, ), )(, ([)], ([(] 

Some sequence of characters ‘(‘, ‘)‘, ‘[‘, and ‘]‘ is given. You are to find the shortest possible regular brackets sequence, that contains the given character sequence as a subsequence. Here, a string a1 a2 ... an is called a subsequence of the string b1 b2 ... bm, if there exist such indices 1 = i1 < i2 < ... < in = m, that aj = bij for all 1 = j = n.
 

Input

The input file contains at most 100 brackets (characters ‘(‘, ‘)‘, ‘[‘ and ‘]‘) that are situated on a single line without any other characters among them.
 

Output

Write to the output file a single line that contains some regular brackets sequence that has the minimal possible length and contains the given sequence as a subsequence.
 

Sample Input

([(]

Sample Output

()[()]

Source

Northeastern Europe 2001
 
解题：这dp啊，我这种学渣啊，每做一次，就有一次新的感觉！水好深啊！
 
注意是单样例的！改成多样例，立马WA了
 
 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib>10 #include <string>11 #include <set>12 #include <stack>13 #define LL long long14 #define pii pair<int,int>15 #define INF 0x3f3f3f3f16 using namespace std;17 const int maxn = 110;18 int dp[maxn][maxn],c[maxn][maxn] = {-1};19 char str[maxn];20 void print(int i,int j) {21     if(i > j) return;22     if(i == j) {23         if(str[i] == ‘(‘ || str[j] == ‘)‘)24             printf("()");25         else printf("[]");26     } else {27         if(c[i][j] >= 0) {28             print(i,c[i][j]);29             print(c[i][j]+1,j);30         } else {31             if(str[i] == ‘(‘) {32                 printf("(");33                 print(i+1,j-1);34                 printf(")");35             } else {36                 printf("[");37                 print(i+1,j-1);38                 printf("]");39             }40         }41     }42 }43 void go() {44     int len = strlen(str),i,j,k,theMin,t;45     for(i = 0; i < len; i++) dp[i][i] = 1;46     for(k = 1; k < len; k++) {47         for(i = 0; i+k < len; i++) {48             j = i+k;49             theMin = dp[i][i]+dp[i+1][j];50             c[i][j] = i;51             for(t = i+1; t < j; t++) {52                 if(dp[i][t]+dp[t+1][j] < theMin) {53                     theMin = dp[i][t]+dp[t+1][j];54                     c[i][j] = t;55                 }56             }57             dp[i][j] = theMin;58             if(str[i] == ‘(‘ && str[j] == ‘)‘ || str[i] == ‘[‘ && str[j] == ‘]‘) {59                 if(dp[i+1][j-1] < theMin) {60                     dp[i][j] = dp[i+1][j-1];61                     c[i][j] = -1;62                 }63             }64         }65     }66     print(0,len-1);67 }68 int main() {69     scanf("%s",str);70     go();71     puts("");72     return 0;73 }
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