题目
poj2955
题意:
给出字符串,求字符串中有(可以不连续)有多少对“ () " 和 “ [] ”
思路:
在区间 [ i , j ] 内,
如果a [ i ] = ‘(’ 并且a [ j ] = ‘)’(或a [ i ] = ‘[’ 并且a [ j ] = ‘]’ ),则
d p [ i ] [ j ] = m a x ( max k = i j − 1 d p [ i ] [ k ] + d p [ k + 1 ] [ j ] , d p [ i + 1 ] [ j − 1 ] ) dp[i][j]=max(\max_{k=i}^{j-1}dp[i][k]+dp[k+1][j], dp[i+1][j-1]) dp[i][j]=max(k=imaxj−1dp[i][k]+dp[k+1][j],dp[i+1][j−1])
否则,
d p [ i ] [ j ] = max k = i j − 1 d p [ i ] [ k ] + d p [ k + 1 ] [ j ] dp[i][j]=\max_{k=i}^{j-1}dp[i][k]+dp[k+1][j] dp[i][j]=k=imaxj−1dp[i][k]+dp[k+1][j]
代码
#include <iostream>
#include <algorithm>
#include <cstring>
#define DEBUG freopen("_in.txt", "r", stdin); freopen("_out1.txt", "w", stdout);
using namespace std;
const int MAXN = 110;
char a[MAXN], b[MAXN], t[] = "([])"; //a:输入的字符串,b:字符串每个字符对应的更改字符
int dp[MAXN][MAXN], need[200];
void solve(){
int lena = strlen(a);
for (int i = 0; i < lena; i++) //记录字符串每个字符对应的字符
b[i] = t[need[a[i]]];
for (int i = 0; i < lena; i++)
dp[i][i] = 0;
for (int len = 1; len < lena; len++)
for (int i = 0; i < lena - len; i++){
int j = i + len;
dp[i][j] = 0;
if (a[i] == b[j] && a[i] < a[j]) //()和[]
dp[i][j] = dp[i+1][j-1] + 2;
for (int k = i; k < j; k++)
dp[i][j] = max(dp[i][j], dp[i][k] + dp[k+1][j]);
}
printf("%d\n", dp[0][lena-1]);
}
int main(){
for (int i = 0; i < 4; i++)
need[t[i]] = 3 - i;
while (~scanf("%s", a) && strcmp(a, "end") != 0)
solve();
return 0;
}