题意：
求区间内 最多正确的括号匹配数。
思路：
令dp[i][j] 表示i~j 区间内 最多正确的括号匹配数。
那么
dp[i][j] = max(dp[i,k] + dp[k+1][j]);
注意边界即可。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 100 + 10;
char s[maxn];
int dp[maxn][maxn];
bool match(int a,int b){
if (a == '(') return b == ')';
if (a == '[') return b == ']';
return 0;
}
int main(){
while(~scanf("%s",s) && s[0] != 'e'){
int len = strlen(s);
for (int i = 0; i < len -1 ; ++i){
dp[i][i] = 0;
if (match(s[i], s[i+1])){
dp[i][i+1] = 2;
}
else dp[i][i+1] = 0;
}
dp[len-1][len-1] = 0;
for (int i = 3; i <= len; ++i){
for (int j = 0; j+i-1 < len; ++j){
int fi = j;
int la = j+i-1;
if (match(s[fi], s[la])){
dp[fi][la] = dp[fi+1][la-1] + 2;
}
else dp[fi][la] = 0;
for (int k = fi; k < la; ++k){
dp[fi][la] = max(dp[fi][la], dp[fi][k] + dp[k+1][la]);
}
}
}
printf("%d\n", dp[0][len-1]);
}
return 0;
}
Brackets Description We give the following inductive definition of a “regular brackets” sequence:
For instance, all of the following character sequences are regular brackets sequences:
while the following character sequences are not:
Given a brackets sequence of characters a1a2 … an, your goal is to find the length of the longest regular brackets sequence that is a subsequence of s. That is, you wish to find the largest m such that for indices i1, i2, …, im where 1 ≤ i1 < i2 < … < im ≤ n, ai1ai2 … aim is a regular brackets sequence. Given the initial sequence Input The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters Output For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line. Sample Input ((()))()()()([]]))[)(([][][)end Sample Output 66406 Source Stanford Local 2004 |
Time Limit: 1000MS | | Memory Limit: 65536K |
Total Submissions: 7916 | | Accepted: 4199 |
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