例题9-10 UVa1626POJ1141 Brackets Sequence(DP)

题意：

看白书

要点：

状态转移方程真难想，这基本上是个区间DP问题，但还得考虑首尾已经是正规的，这两种情况都要考虑，而且还互相干扰，最后还得打印解，基本就是重新检查一下哪个决策最好。还有个陷阱：输入串可能是空串，所以要用gets。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
using namespace std;
char str[1000];
int d[1000][1000];
int n;bool match(char a, char b)
{if ((a == '('&&b == ')')||(a == '['&&b == ']'))return true;return false;
}
void dp()
{int i, j, k, l;memset(d, 0, sizeof(d));for (i = 0; i <= n; i++)d[i][i] = 1;		//单字符为0for (l = 2; l <= n; l++){for (i = 0; i + l <= n; i++){j = i + l-1;d[i][j] = 0x3f3f3f3f;if (match(str[i],str[j]))d[i][j] = min(d[i][j],d[i + 1][j - 1]);for (k = i; k <= j-1; k++)d[i][j] = min(d[i][j], d[i][k] + d[k+1][j]);}}
}
void print(int i, int j)
{if (i > j)return;if (i == j){if (str[i] == '(' || str[i] == ')')printf("()");elseprintf("[]");return;}int ans = d[i][j];if (match(str[i], str[j]) && (ans == d[i + 1][j - 1])){printf("%c", str[i]); print(i + 1, j - 1); printf("%c", str[j]);return;}for (int k = i; k <= j - 1; k++){if (ans == d[i][k] + d[k + 1][j]){print(i, k);print(k + 1, j);return;}}
}int main()
{gets_s(str);n = strlen(str);dp();print(0, n-1);printf("\n");return 0;
}