标签:print sharp ++ main 数组 bit stdin txt inpu
题目链接:
https://cn.vjudge.net/contest/313499#problem/L
SOLUTION:
对于题意,要求最短的表达式,当用最小循环元来表示一个字符串时,其表达式才最短。
对于字符串S自匹配求出next数组,分析可以发现:当i-next[i]能整除i时,S[1~i-next[i]]就是S[1~i]的最小循环元。它的最大循环次数就是i/(i-next[i])。
接下来枚举所有字串的最小循环元,取最优。
CODE:
#include "bits/stdc++.h"
using namespace std;
const int maxn = 1e4 + 100;
int n;
char s[maxn];
int Next[maxn];
int f[maxn];
void getnext(char str[], int l) {
for (int i = 2, j = 0; i <= l; i++) {
while (j > 0 && str[i] != str[j + 1]) j = Next[j];
if (str[i] == str[j + 1]) j++;
Next[i] = j;
}
}
int main() {
//freopen("input.txt", "r", stdin);
int N, now, temp;
scanf("%d", &N);
while (N--) {
scanf("%s", s + 1);
n = strlen(s + 1);
for (int i = 0; i <= n; i++)
f[i] = i;
for (int i = 1; i <= n; i++) {
getnext(s + i - 1, n - i + 1);
for (int j = i; j <= n; j++) {
f[j]=min(f[j],j-i+1+f[i-1]);
now = j - i + 1;
if (now % (now - Next[now]) == 0) {
f[j] = min(f[j], f[i - 1] + now - Next[now]);
}
}
}
printf("%d\n", f[n]);
}
return 0;
}
标签:print sharp ++ main 数组 bit stdin txt inpu
原文地址:https://www.cnblogs.com/zhangbuang/p/11279023.html