POJ - 2955 Brackets

题意:给一个有'(' ')' '[' ']'组成的字符串,问满足括号匹配的最大长度子串。

思路:区间dp,dp[i][j]代表区间i到j的ans,若i与j匹配,dp[i][j]=dp[i+1][j-1]+2,不匹配则枚举断点,dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j])

点击打开链接

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char s[102];
int dp[102][102];
int main(){
    int i, j, k, x;
    while(gets(s)!=NULL){
        if(s[0]=='e')break;
        memset(dp,0,sizeof(dp));
        int len= strlen(s);
        for(k=1;k<len;k++) {
            for(i=0,j=k;j<len;i++,j++){
                if(s[i]=='('&&s[j]==')'||s[i]=='['&&s[j]==']')
                    dp[i][j]=dp[i+1][j-1]+2;
                for(x=i;x<j;x++) 
                    dp[i][j]=max(dp[i][j],dp[i][x]+dp[x+1][j]);
            }
        }
        printf("%d\n",dp[0][len-1]);
    }
    return 0;
}

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转载自blog.csdn.net/qq_41231363/article/details/80146596