Brackets sequence UVA
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区间DP
题目描述:输出最短的满足括号匹配的字符串,该字符串满足,输入时该字符串的子串。
解题分析:先做的POJ 2955-Brackets,两道题虽然不一样,但是该题的想法可以应用到本题上,定义dp[i][j]为以第i个字符开头,以第j个字符结尾的,最长匹配子串的长度。因为对于输入串最长的满足括号匹配的子串是最长的不需要添加的子串,可以直接输出,不匹配的才需要添加。利用POJ 2955-Brackets的状态转移方程,把打印函数改一改就行,注意空串的问题,神坑啊,找了半天才发觉这个问题。
代码如下:
#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int maxn = 200 + 10;int dp[maxn][maxn];char s[maxn];int len;void out(int u,int v)//dp[i][j] = 0,意味着这个子串没有匹配的括号,直接添加并输出即可。{ for(int i = u; i <= v; i++) { if(s[i] == '(' || s[i] == ')') { printf("()"); } else if(s[i] == '[' || s[i] == ']') { printf("[]"); } }}void Print(int u,int v){ if(u > v) return; if(u == v) { out(u,v); return; } if(dp[u + 1][v - 1] + 2 == dp[u][v] && (s[u] == '(' && s[v] ==')' || s[u] == '[' && s[v] ==']' )) { printf("%c",s[u]); Print(u + 1,v - 1); printf("%c",s[v]); } else { for(int k = u; k < v; k++) { if(dp[u][v] == dp[u][k] + dp[k+1][v]) { if(dp[u][k] == 0 && dp[k + 1][v] == 0) { out(u,k); out(k + 1,v); } else if(dp[u][k] == 0 && dp[k+1][v] != 0) { out(u,k); Print(k + 1,v); } else if(dp[u][k] != 0 && dp[k+1][v] == 0) { Print(u,k); out(k + 1,v); } else { Print(u,k); Print(k + 1,v); } break; } } }}int main(){ int tt; int kase = 0; scanf("%d",&tt); getchar(); getchar(); while(tt--) { if(kase) getchar(); gets(s + 1); int n = strlen(s + 1); memset(dp,0,sizeof(dp)); for(int i = n; i >= 1; i--) { for(int j = i + 1; j <= n; j++) { for(int k = i; k < j; k++) { dp[i][j] = max(dp[i][j],dp[i][k] + dp[k+1][j]); } if(s[i] == '(' && s[j] ==')' || s[i] == '[' && s[j] ==']') { dp[i][j] = max(dp[i][j],dp[i + 1][j - 1] + 2); } } } if(kase++) printf("\n"); Print(1,n); printf("\n"); } return 0;}
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