poj2955（括号匹配问题）

来源：互联网发布：仿购物商城 app 源码编辑：程序博客网时间：2024/06/05 20:48

题意：给你一串()[]括号，要你求出这串括号的最大匹配个数，如'('与')'匹配，为2个，'['与']'匹配，为2个，其他不能匹配.......

思路：dp[i][j]代表从区间i到区间j所匹配的括号的最大个数，首先，假设不匹配，那么dp[i][j]=dp[i+1][j]；然后查找i+1~~j有木有与第i个括号匹配的

有的话，dp[i][j]=max(dp[i][j],dp[i+1][k-1]+dp[k][j]+2).....

代码：

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#define INF 0x3fffffffusing namespace std;char a[105];int dp[105][105];int main(){    char b[6] = "end";    while(scanf("%s",a) != EOF){        if(!strcmp(b,a))break;        int n = strlen(a);        memset(dp,0,sizeof(dp));        for(int i = 2; i <= n; i++){            for(int j = 0; j < n; j++){                int theend = j+i-1;                dp[j][theend] = dp[j][theend-1];                if(a[theend] == ')'){                    for(int k = theend-1; k >= j; k--){                        if(a[k] == '(')                            dp[j][theend] = max(dp[j][theend],dp[j][k-1]+dp[k+1][theend-1]+2);                    }                }                if(a[theend] == ']'){                    for(int k = theend-1; k >= j; k--){                        if(a[k] == '[')                            dp[j][theend] = max(dp[j][theend],dp[j][k-1]+dp[k+1][theend-1]+2);                    }                }            }        }        cout << dp[0][n-1] << endl;    }}

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