HDU3613 Best Reward

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=3613

题意：多组数据,给定每个字母的价值和一个串S，要把这个串S分成两个串T1、T2，若某串T是回文串那么就能获得该串上字母的价值，否则可获得的价值为0，求最大价值

题解：RT 用exkmp或者马拉车搞一搞就好了

心得什么的：撒比的我想着用exkmp搞，练习一下，结果..一搞就搞了半个世纪qwq至今WA原因不明。于是气愤的我改用Manacher水过了= =

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
#define maxn 510000

int len,n,w[maxn],a[maxn],g[maxn*2];
char s[maxn],ss[maxn*2];
int mymin(int x,int y){return (x<y)?x:y;}
int mymax(int x,int y){return (x>y)?x:y;}
void init()
{
	int i;len=0;
	for (i=1;i<=26;i++) scanf("%d",&a[i]);scanf("
");
	gets(s+1);n=strlen(s+1);w[0]=0;
	for (i=1;i<=n;i++) w[i]=w[i-1]+a[s[i]-'a'+1];
	ss[len]='|';
	for (i=1;i<=n;i++)
	{
		ss[++len]=s[i];
		ss[++len]='|';
	}
}
void Manacher()
{
	int i,mx=0,p=0;
	memset(g,0,sizeof(g));
	g[0]=1;
	for (i=1;i<=len;i++)
	{
		if (i<mx) g[i]=mymin(g[p*2-i],mx-i);
		else g[i]=1;
		while (i+g[i]<=len && ss[i+g[i]]==ss[i-g[i]]) g[i]++;
		if (i+g[i]>mx) mx=i+g[i],p=i;
	}
}
void ges()
{
	int i,ans=0,sum;
	for (i=1;i<n;i++)
	{
		sum=0;
		if (g[i]-1==i) sum+=w[i];//前i个是回文
		int xx=i+n;
		if (n-i==g[xx]-1) sum+=w[n]-w[i];//后n-i+1个是回文
		ans=mymax(ans,sum);
	}printf("%d
",ans);
}
int main()
{
	//freopen("a.in","r",stdin);
	//freopen("a.out","w",stdout);
	int T;
	scanf("%d",&T);
	while (T--)
	{
		init();
		Manacher();
		ges();
	}
	return 0;
}

还有WA的exkmp的代码，，，ORZ有没有好心的大大帮我看看那里有问题【我的做法挺蠢的qwq

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 510000

int n,w[maxn],a[30];
int ex[3][maxn],nt[3][maxn];
char s[3][maxn];
int mymax(int x,int y){return (x>y)?x:y;}
int mymin(int x,int y){return (x<y)?x:y;}
void exkmp(int l1,int l2,int tp)
{
	int id,mx,i;
	memset(nt[tp],0,sizeof(nt[tp]));
	id=1;mx=0;nt[tp][1]=n;
	for (i=2;i<=n;i++)
	{
		if (i+nt[tp][i-id+1]-1<mx) nt[tp][i]=nt[tp][i-id+1];
		else
		{
			nt[tp][i]=mymax(0,mx-i+1);
			while (i+nt[tp][i]<=n && s[l1][i+nt[tp][i]]==s[l1][1+nt[tp][i]]) nt[tp][i]++;
			if (i+nt[tp][i]-1>mx) mx=i+nt[tp][i]-1,id=i;
		}
	}
		
	memset(ex[tp],0,sizeof(ex[tp]));
	id=0;mx=0;
	for (i=1;i<=n;i++)
	{
		if (i+nt[tp][i-id+1]-1<mx) ex[tp][i]=nt[tp][i-id+1];
		else
		{
			ex[tp][i]=mymax(0,mx-i+1);
			while (i+ex[tp][i]<=n && s[l1][1+ex[tp][i]]==s[l2][i+ex[tp][i]]) ex[tp][i]++;
			if (i+ex[tp][i]-1>mx) mx=i+ex[tp][i]-1,id=i;
		}
	}
}
int main()
{
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
	int T,i,ans,sum,dz;
	scanf("%d",&T);
	while (T--)
	{
		for (i=1;i<=26;i++) scanf("%d",&a[i]);scanf("
");
		gets(s[1]+1);n=strlen(s[1]+1);w[0]=0;
		for (i=1;i<=n;i++) s[2][n-i+1]=s[1][i];
		for (i=1;i<=n;i++) w[i]=w[i-1]+a[s[1][i]-'a'+1];
		exkmp(1,2,1);exkmp(2,1,2);ans=0;
		if (n==2) ans=w[n];
		for (i=2;i<n;i++)
		{
			sum=0;
			if (ex[1][n-i+1]==i) sum+=w[i];//表示前i个字符是回文串
			if (ex[2][i+1]==n-i) sum+=w[n]-w[i];//表示后n-i个字符是回文串
			ans=mymax(ans,sum);
		}printf("%d
",ans);
	}
	return 0;
}