Ural 1635. Mnemonics and Palindromes

pasoib

6
p a s o i b

zzzqxx

3
zzz q xx

wasitacatisaw

1
wasitacatisaw

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define pi acos(-1.0)
#define maxn (4000 + 50)
#define mol 1000000009
#define inf 0x3f3f3f3f
#define Lowbit(x) (x & (-x))
using namespace std;
typedef long long int LLI;

//dp[i]表示从0到i最少有多少回文子串，初始为最大值
//dp[i] = min(dp[j] + 1);
//pre[i]存的是以i结尾的回文子串的首地址(下标)

char str[maxn];
int dp[maxn];
int vis[maxn][maxn];
int pre[maxn];

void handle(int n) {
//vis[i][j]表示从i到j是不是回文串
//依据原理：如果i到j不是回文的，那么i-1到j+1也不是回文的
for(int i = 0; i < n; i ++) {
for(int j = 0;; j ++) {//这是以i为对称轴，左右分别延伸j长度的
if(i - j < 0)   break;
if(i + j >= n)   break;
if(str[i + j] != str[i - j])    break;
vis[i - j][i + j] = true;
}
for(int j = 1;; j ++) {//这是以i和i+1中间为对称轴
if(i - j + 1 < 0)   break;
if(i + j >= n)      break;
if(str[i - j + 1] != str[i + j])    break;
vis[i - j + 1][i + j] = true;
}
}
}
stack > asta;
int main() {
//    freopen("in.txt", "r" , stdin);
//    freopen("out.txt", "w" , stdout);
scanf("%s",str);
int len = strlen(str);
fill(dp,dp + maxn,inf);
fill(pre,pre + maxn,-1);
memset(vis,false,sizeof(vis));
handle(len);
dp[0] = 1;
for(int i = 0; i < len; i ++) {
if(vis[0][i])   dp[i] = 1,pre[i] = 0;
}
for(int i = 0;i < len;i ++){
if(dp[i] == 1)  continue;
for(int j = 0;j < i;j ++){
if(vis[j + 1][i] && dp[i] > dp[j] + 1){
dp[i] = dp[j] + 1;
pre[i] = j + 1;
}
}
}
printf("%d\n",dp[len - 1]);
int p = pre[len - 1],e = len - 1;
while(e != -1 && p != -1){
asta.push(pair(p,e));
e = p - 1;
p = pre[e];
}
while(!asta.empty()){
int x = asta.top().first;
int y = asta.top().second;
asta.pop();
for(int i = x;i <= y;i ++){
printf("%c",str[i]);
}
if(!asta.empty())   printf(" ");
}
return 0;
}