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POJ #1141 - Brackets Sequence - TODO: POJ website issue
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A bottom-up DP. To be honest, it is not easy to relate DP to this problem. Maybe, all "most"\"least" problems can be solved using DP..
Reference: http://blog.sina.com.cn/s/blog_8e6023de01014ptz.html
There's an important details to AC: in case of "())", there are 2 solutions: ()() and (()). For the AC code, the former one is preferred.
// 1141
// http://blog.sina.com.cn/s/blog_8e6023de01014ptz.html
//
#include <stdio.h>
#include <string.h>
#include <memory.h> #define MAX_LEN 101 bool isMatched(char *in, int i, int j)
{
return (in[i] == '(' && in[j] == ')') || (in[i] == '[' && in[j] == ']');
}
void printPath(int path[MAX_LEN][MAX_LEN], int i, int j, char in[MAX_LEN])
{
int sInx = path[i][j];
if (sInx == -)
{
if (i == j)
{
//printf("Complete @ %d\n", i);
switch (in[i])
{
case '(':
case ')': printf("()"); break;
case '[':
case ']': printf("[]"); break;
}
return;
}
else if (i + == j)
{
//printf("Already matched: [%d, %d]\n", i, j);
printf("%c%c", in[i], in[j]);
return;
}
else if ((i+) < j)
{
printf("%c", in[i]);
printPath(path, i + , j - , in);
printf("%c", in[j]);
}
}
else
{
printPath(path, , path[i][j], in);
//printf("Break @ %d\n", path[i][j], in);
printPath(path, path[i][j] + , j, in);
}
} void calc(char in[MAX_LEN])
{
unsigned slen = strlen(in);
if (slen == )
{
printf("\n");
return;
}
else if (slen > )
{
int dp[MAX_LEN][MAX_LEN];
int path[MAX_LEN][MAX_LEN]; // Init
for (int i = ; i < MAX_LEN; i ++)
for (int j = ; j < MAX_LEN; j++)
{
dp[i][j] = 0xFFFFFF;
path[i][j] = -;
} // Def: dp[i][j] = min num of chars to add, from i to j
// Recurrence relation:
// 1. dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j]), where k in (i..j)
// 2. dp[i][j] = dp[i+1][j-1], <IF in[i]:in[j] = [] or ()> // i..j is range, k is interval
for (int k = ; k < slen; k++) // NOTE: this is a bottom-up DP. We have to go from 0 to slen as interval
for (int i = , j = i + k; i < slen - k; i ++, j ++)
{
if (i == j)
{
dp[i][j] = ;
path[i][j] = -;
continue;
}
bool bIsMatched = isMatched(in, i, j);
if (bIsMatched) // eq 2
{
if (k == ) // length == 2
{
dp[i][j] = ;
path[i][j] = -;
continue;
}
else if (k > ) // length > 2
{
dp[i][j] = dp[i + ][j - ];
path[i][j] = -; // we don't split matched pair
// A: we still go ahead with eq1
}
}
//else // eq1
{
// t is iterator of split index
for (int t = i; t < j; t++)
{
int newVal = dp[i][t] + dp[t + ][j];
if (newVal <= dp[i][j]) // Label A: we prefer splitted solution
{
dp[i][j] = newVal;
path[i][j] = t;
}
}
}
}
printPath(path, , slen - , in);
} // if (slen > 0)
} int main()
{
char in[MAX_LEN] = { };
gets(in);
calc(in);
printf("\n");
return ;
}
