uva1626 poj 1141 Brackets Sequence 区间dp 打印路径

// poj 1141 Brackets Sequence// 也是在紫书上看的一题，uva就是多了一个t组数据。// 经典区间dp// dp(i,j)表示区间[i,j]内所需要增加的括号数目// 则分为两种情况// 一种是s[i]和s[j]是匹配的则// dp[i][j] = min(dp[i][j],dp[i+1][j-1])// 另外一种情况是不匹配// dp[i][j] = min(dp[i][j],dp[i][k]+dp[k+1][j]){i<k<j};// 但是无论如何都要进行第二种情况// 比如[][]这种情况//// 这题其实还是挺简单的，但是就是打印的方法可能有点复杂// 递归打印果然是神奇的东西啊//// 顺便说一句，数据有点坑，有空串的情况，我因此wa了将近一个小时//// 学到了递归打印，简单的区间dp//// 哎，继续练吧。。。。#include <algorithm>#include <bitset>#include <cassert>#include <cctype>#include <cfloat>#include <climits>#include <cmath>#include <complex>#include <cstdio>#include <cstdlib>#include <cstring>#include <ctime>#include <deque>#include <functional>#include <iostream>#include <list>#include <map>#include <numeric>#include <queue>#include <set>#include <stack>#include <vector>#define ceil(a,b) (((a)+(b)-1)/(b))#define endl '\n'#define gcd __gcd#define highBit(x) (1ULL<<(63-__builtin_clzll(x)))#define popCount __builtin_popcountlltypedef long long ll;using namespace std;const int MOD = 1000000007;const long double PI = acos(-1.L);template<class T> inline T lcm(const T& a, const T& b) { return a/gcd(a, b)*b; }template<class T> inline T lowBit(const T& x) { return x&-x; }template<class T> inline T maximize(T& a, const T& b) { return a=a<b?b:a; }template<class T> inline T minimize(T& a, const T& b) { return a=a<b?a:b; }const int maxn = 1008;int d[maxn][maxn];char s[maxn];int n;const int inf = 0x4f4f4f4f;bool match(char a,char b){if (a=='(' && b==')')return true;if (a=='[' && b==']')return true;return false;}int dp(int i,int j){if (i==j)return d[i][j];if (i>j)return d[i][j] = 0;if (d[i][j]!=inf)return d[i][j];int& ans = d[i][j];if (match(s[i],s[j]))ans = min(ans,dp(i+1,j-1));for (int k=i;k<j;k++)ans = min(ans,dp(i,k)+dp(k+1,j));return ans;}void print(int i,int j){if (i>j)return ;if (i==j){if (s[i]=='('||s[i]==')')printf("()");else printf("[]");return ;}int ans = d[i][j];if (match(s[i],s[j]) && ans == d[i+1][j-1]){printf("%c",s[i]);print(i+1,j-1);printf("%c",s[j]);return ;}for (int k=i;k<j;k++){if (ans == d[i][k]+d[k+1][j]){print(i,k);print(k+1,j);return ;}}}void init(){n = strlen(s);for (int i=0;i<=n;i++)for (int j=0;j<=n;j++)d[i][j] = inf;for (int i=0;i<=n;i++){d[i][i] = 1;}dp(0,n-1);print(0,n-1);puts("");}int main() {//freopen("G:\\Code\\1.txt","r",stdin);while(gets(s)){    if (s[0])            init();        else puts("");}return 0;}