UVA 11584 Partitioning by Palindromes 划分成回文串(DP + 预处理)
2016-09-30 08:30
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大大体题意:
给你n个字符串,求出能把这个字符串划分成最少几个回文串?
思路:
很简单的dp,做了好几遍了,今天才优化到n^2的复杂度= =!
令dp[i],表示从位置1到位置i 最少划分的回文串数!
那么直接二重循环,如果j~i是回文串的化,那么dp[i] = min(dp[i],dp[j-1]+1);
判断j~i是否回文直接预处理即可!
方法时枚举回文串的中心,分奇数偶数讨论以下, 当两段字符不相同跳出循环,否则记录!
以下代码是10ms代码,还待优化!
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1000 + 10;
char s[maxn];
int dp[maxn];
bool g[maxn][maxn];
int main(){
int T;
scanf("%d",&T);
while(T--){
memset(g,0,sizeof g);
memset(dp,0x3f3f3f3f,sizeof dp);
dp[0] = 0;
scanf("%s",s+1);
int len = strlen(s+1);
for (int i = 1; i <= len; ++i){
g[i][i] = 1;
for (int j = 1; ; ++j){
if (i-j < 1 || i+j > len)break;
if (s[i-j] == s[i+j])g[i-j][i+j] = 1;
else break;
}
if (s[i] == s[i+1]){
g[i][i+1] = 1;
for (int j = 1; ; ++j){
if (i-j < 1 || i+1+j > len)break;
if (s[i-j] != s[i+1+j])break;
g[i-j][i+1+j] = 1;
}
}
}
for (int i = 1; i <= len; ++i){
for (int j = 1; j <= i; ++j){
if (g[j][i]) dp[i] = min(dp[j-1] + 1,dp[i]);
}
}
printf("%d\n",dp[len]);
}
return 0;
}
给你n个字符串,求出能把这个字符串划分成最少几个回文串?
思路:
很简单的dp,做了好几遍了,今天才优化到n^2的复杂度= =!
令dp[i],表示从位置1到位置i 最少划分的回文串数!
那么直接二重循环,如果j~i是回文串的化,那么dp[i] = min(dp[i],dp[j-1]+1);
判断j~i是否回文直接预处理即可!
方法时枚举回文串的中心,分奇数偶数讨论以下, 当两段字符不相同跳出循环,否则记录!
以下代码是10ms代码,还待优化!
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1000 + 10;
char s[maxn];
int dp[maxn];
bool g[maxn][maxn];
int main(){
int T;
scanf("%d",&T);
while(T--){
memset(g,0,sizeof g);
memset(dp,0x3f3f3f3f,sizeof dp);
dp[0] = 0;
scanf("%s",s+1);
int len = strlen(s+1);
for (int i = 1; i <= len; ++i){
g[i][i] = 1;
for (int j = 1; ; ++j){
if (i-j < 1 || i+j > len)break;
if (s[i-j] == s[i+j])g[i-j][i+j] = 1;
else break;
}
if (s[i] == s[i+1]){
g[i][i+1] = 1;
for (int j = 1; ; ++j){
if (i-j < 1 || i+1+j > len)break;
if (s[i-j] != s[i+1+j])break;
g[i-j][i+1+j] = 1;
}
}
}
for (int i = 1; i <= len; ++i){
for (int j = 1; j <= i; ++j){
if (g[j][i]) dp[i] = min(dp[j-1] + 1,dp[i]);
}
}
printf("%d\n",dp[len]);
}
return 0;
}
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