NYOJ 括号匹配系列2，5

本文出自：http://blog.csdn.net/svitter

括号匹配一：http://acm.nyist.net/JudgeOnline/problem.php?pid=2

括号匹配二：http://acm.nyist.net/JudgeOnline/problem.php?pid=15

之前被这个题目难住，现在看动态规划就顺便过来AC了它。结果发现当年被难住一点也不丢人。。

括号匹配一很简单，就是栈的应用，AC代码：

//============================================================================// Name        : 括号匹配.cpp// Author      : // Version     :// Copyright   : Your copyright notice// Description : Hello World in C++, Ansi-style//============================================================================#include <iostream>#include <cstdio>#include <string.h>#include <stack>using namespace std;void ace(){int n;scanf("%d", &n);char ch;char tmp;ch = getchar();while(n --){stack <char> s;while((ch = getchar())!= '\n'){if(s.empty())s.push(ch);else{tmp = s.top();if(tmp == '(' && ch == ')')s.pop();else if(tmp == '[' && ch == ']')s.pop();elses.push(ch);}}if(s.empty())printf("Yes\n");elseprintf("No\n");}}int main() {ace();return 0;}

第二道就是DP题目了- -

真心被难住了。下面分析一下：

通过分析（别问我怎么分析的，画多了就看出来了- -）这必定是一个通过区间括号求和计算出的最小匹配括号值。

dp方程： dp [ i ] [ j ] = min ( dp [ i ] [ j ] , dp [ i ] [ k ] + dp [ k + 1 ] [ j ] );

dp[ i ][ j ] 表示当前匹配最小的括号值。后来发现这个不是正确的- -。因为这个阶段值与另一个阶段值会相互影响，违反了条件。

有重新做了分析：

发现无非就是这么几种情况：

" ..[ ... ] " + " ] “

" ..[ ... [ " + " ] "

" ..[ ... ] " + " [ "

" ..[ ... [ " + " [ "

这么四种情况。

如果假设dp [ i ] [ j ] = dp [ i ] [ j - 1 ] + 1

那么不符合情况的有第一种和第二种。而这两种情况就是因为中间串中有能够与最新加入的str[j]匹配的串。所以，当出现匹配串时，寻找最佳的匹配方案 ——dp [ i ] [ j ] = min ( dp [ i ] [ j ] , dp [ i ] [ k - 1 ] + dp [ k + 1 ] [ j - 1 ] );就是去除了两个括号，求括号里面的部分和括号外面部分的最小值。

特别的，为了针对 j  == i + 1的情况， dp [ i ] [ j ] = min ( dp [ i ] [ j ], dp [ i + 1] [ k - 1 ] + dp [ k + 1 ] [ j ])是不成立的。

AC代码：

//============================================================================// Name        : 括号匹配.cpp// Author      :// Version     :// Copyright   : Your copyright notice// Description : Hello World in C++, Ansi-style//============================================================================#include <iostream>#include <cstdio>#include <string.h>#include <stack>using namespace std;#define min(a, b) a > b ? b : aint dp[102][102];char str[1001];bool match(int i, int j){    if (str[i] == '(' && str[j] == ')')        return true;    else if (str[i] == '[' && str[j] == ']')        return true;    else        return false;}void ace(){    //case    int c;    scanf("%d", &c);    getchar();    //work point    int i, j, k;    //value    int n;    while (c--)    {        scanf("%s", str + 1); //此处可以尝试a+1        memset(dp, 0, sizeof(dp));        n = strlen(str + 1);        //区间为差值为0时，必定需要一个括号匹配        for (i = 1; i <= n; i++)            dp[i][i] = 1;        for (j = 2; j <= n; j++)        // j = 2...n            for (i = j - 1; i >= 1; i--) // i = j...1            {                dp[i][j] = dp[i][j-1] + 1;                for (k = i; k < j; k++)  //k = i+1...j-1                {                    if(match(k, j))                    {                        dp[i][j] = min(dp[i][j], dp[i][k-1] + dp[k + 1][j - 1]);                    }                }            }        printf("%d\n", dp[1][n]);    }}int main(){    ace();    return 0;}

后来依据http://blog.csdn.net/svitter/article/details/25186367

重写了代码，解题思路可以看上述题目。

//============================================================================// Name        : test.cpp// Author      : // Version     :// Copyright   : Your copyright notice// Description : Hello World in C++, Ansi-style//============================================================================//============================================================================// Name        : 动态规划.cpp// Author      : blog.csdn.net/svitter// Version     :// Copyright   : Your copyright notice// Description : Hello World in C++, Ansi-style//============================================================================#include <iostream>#include <stdio.h>#include <string.h>using namespace std;#define MAXN 256char br[MAXN];int dp[MAXN][MAXN], pos[MAXN][MAXN];int len;bool match(int i, int j) {if (br[i] == '(' && br[j] == ')')return true;if (br[i] == '[' && br[j] == ']')return true;return false;}int main() {//work pitint i, j, k, mid, t;int Case;scanf("%d", &Case);getchar();while (Case--) {while (gets(br) != NULL) {memset(dp, 0, sizeof(dp));len = strlen(br);for (i = 0; i < len; i++)dp[i][i] = 1;for (k = 1; k < len; k++) {for (i = 0; i + k < len; i++) {j = i + k;dp[i][j] = 0x7fffffff;if (match(i, j)) { //如果当前位置匹配，那么pos置-1dp[i][j] = dp[i + 1][j - 1], pos[i][j] = -1;}for (mid = i; mid < j; mid++) {if (dp[i][j] > (t = dp[i][mid] + dp[mid + 1][j])) { //如果存在更优分解，那么选择更优分解dp[i][j] = t, pos[i][j] = mid;}}}}printf("%d\n", dp[0][len - 1]);}}return 0;}