题意:问最多有几个括号匹配
思路:用dp[i][j]表示i到j最多匹配,若i和j构成匹配,那么dp[i][j] = dp[i + 1][j - 1] + 2,剩下情况dp[i][j] = max(dp[i][j], dp[i][k] + dp[k + 1][j])
代码:
#include<set> #include<map> #include<cmath> #include<queue> #include<cstdio> #include<cstring> #include<algorithm> typedef long long ll; using namespace std; const int maxn = 100 + 10; const int MOD = 1e9 + 7; const int INF = 0x3f3f3f3f; int dp[maxn][maxn]; char s[maxn]; bool ok(int i, int j){ if(s[i] == ‘(‘ && s[j] == ‘)‘) return true; if(s[i] == ‘[‘ && s[j] == ‘]‘) return true; return false; } int main(){ while(~scanf("%s", s + 1)){ if(strcmp(s + 1, "end") == 0) break; memset(dp, 0, sizeof(dp)); int n = strlen(s + 1); for(int len = 2; len <= n; len++){ for(int i = 1; i + len - 1 <= n; i++){ int j = i + len - 1; if(ok(i, j)){ dp[i][j] = dp[i + 1][j - 1] + 2; } for(int k = i; k < j; k++){ dp[i][j] = max(dp[i][j], dp[i][k] + dp[k + 1][j]); } } } printf("%d\n", dp[1][n]); } return 0; }
原文地址:https://www.cnblogs.com/KirinSB/p/10317127.html
时间: 2024-11-06 20:45:14