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 0x3fffffff using 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; } }
