ICPC Central Europe Regional Contest 2019 A. ABB（馬拉車，哈希）

Fernando was hired by the University of Waterloo to finish a development project the university started some time ago. Outside the campus, the university wanted to build its representative bungalow street for important foreign visitors and collaborators.

Currently, the street is built only partially, it begins at the lake shore and continues into the forests, where it currently ends. Fernando’s task is to complete the street at its forest end by building more bungalows there. All existing bungalows stand on one side of the street and the new ones should be built on the same side. The bungalows are of various types and painted invarious colors.

The whole disposition of the street looks a bit chaotic to Fernando. He is afraid that it will look even more chaotic when he adds new bungalows of his own design. To counter balance the chaosof all bungalow shapes, he wants to add some order to the arrangement by choosing suitablecolors for the new bungalows. When the project is finished, the whole sequence of bungalowcolors will be symmetric, that is, the sequence of colors is the same when observed from eitherend of the street.

Among other questions, Fernando wonders what is the minimum number of new bungalows heneeds to build and paint appropriately to complete the project while respecting his self-imposed bungalow color constraint.

Input Specification

The first line contains one integer N (1 \leq N \leq 4·10^{5})N(1≤N≤4⋅10
5
), the number of existing bungalows in the street. The next line describes the sequence of colors of the existing bungalows, from the beginning of the street at the lake. The line contains one string composed of N lowercase letters(“a” through “z”), where different letters represent different colors.

Output Specification

Output the minimum number of bungalows which must be added to the forest end of the street and painted appropriately to satisfy Fernando’s color symmetry demand.

樣例輸入1複製
3
abb
樣例輸出1複製
1
樣例輸入2複製
12
recakjenecep
樣例輸出2複製
11

題意：
在字符串後面添加一些字符使得其稱爲迴文串。

思路：
只需要找到最大的迴文串後綴即可。
比賽的時候用的馬拉車，但是哈希也完全ok。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 1e6 + 7;
char s1[maxn],s2[maxn];
int f[maxn];

int manacher(int len)
{
    int ans = 0,id = 1,mx = 0;
    for(int i = 1;i < len;i++)
    {
        if(id + mx > i)f[i] = min(f[id * 2 - i],id + mx - i);
        while(i - f[i] - 1 >= 1 && i + f[i] + 1 <= len && s2[i - f[i] - 1] == s2[i + f[i] + 1])
            f[i]++;
        if(id + mx < i + f[i])
        {
            id = i;mx = f[i];
        }
        ans = max(ans,f[i]);
    }
    return ans;
}

int main()
{
    int n;scanf("%d",&n);
    scanf("%s",s1 + 1);
    int len = (int)strlen(s1 + 1);
    int len2 = 0;
    for(int i = 1;i <= len;i++)
    {
        s2[++len2] = '#';
        s2[++len2] = s1[i];
    }
    s2[++len2] = '#';
    manacher(len2);
    int ans = 0;
    for(int i = 1;i <= len2;i++) {
        if(f[i] >= len2 - i) {
            ans = max(ans,f[i]);
        }
    }

    printf("%d\n",n - ans);
    return 0;
}

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

typedef unsigned long long ull;

const int maxn = 4e5 + 7;
const int mod = 1e8 + 7;
char s[maxn];

ull h1[maxn],h2[maxn],jie[maxn];

bool check1(int x,int y) { //以x爲中心
    ull H1 = h1[x - 1] - h1[x - 1 - y] * jie[y];
    ull H2 = h2[x + 1] - h2[x + y + 1] * jie[y];
    return H1 == H2;
}

bool check2(int x,int y) { //x爲右半部分左端點
    ull H1 = h1[x - 1] - h1[x - 1 - y] * jie[y];
    ull H2 = h2[x] - h2[x + y] * jie[y];
    return H1 == H2;
}

int main() {
    int n;scanf("%d",&n);
    scanf("%s",s + 1);
    jie[0] = 1;
    for(int i = 1;i <= n;i++) {
        jie[i] = jie[i - 1] * mod;
        h1[i] = h1[i - 1] * mod + s[i];
    }
    for(int i = n;i >= 1;i--) {
        h2[i] = h2[i + 1] * mod + s[i];
    }
    
    int ans = 0;
    for(int i = n;i >= n / 2;i--) {
        if(i - 1 >= n - i && check1(i,n - i)) {
            ans = max(2 * (n - i) + 1,ans);
        }
        if(i - 1 >= n - i + 1 && check2(i,n - i + 1)) {
            ans = max(2 * (n - i + 1),ans);
        }
    }
    printf("%d\n",n - ans);
    return 0;
}