2015
04-14

# hdu 3973-ac’s string-字符串-[解题报告]hoj

http://acm.hdu.edu.cn/showproblem.php?pid=3973

Problem Description
You are given some words {Wi}. Then our stupid AC will give you a very long string S. AC is stupid and always wants to know whether one substring from S exists in the given words {Wi} .

For example, S = "abcd", and the given words {Wi} = {"bc", "ad", "dd"}. Then Only S[2..3] = "bc" exists in the given words. (In this problem, the first element of S has the index "0".)

However, this is toooooooooooo easy for acmers ! The stupid and evil AC will now change some letters in S. So could you solve this problem now?

Input
The first line is one integer T indicates the number of the test cases. (T <=20)

Then for every case, there is one integer n in the first line indicates the number of the given words(The size of the {Wi}) . Then n lines has one string which only has ‘a’- ‘z’. (1 <= n <= 10000, sigma|Wi| <= 2000000) .

Then one line has one string S, here |S| <= 100000.

Then one integer m, indicating the number of operations. (1 <= m <= 100000)

Then m lines , each line is the operation:

(1)Q L R , tell AC whether the S[L..R] exists in the given strings ;

(2)C X Y , chang S[X] to Y, here Y : ‘a’-'z’ .

Output
First output “Case #idx:” in a single line, here idx is the case number count from 1.Then for each "Q" operation, output "Yes" if S[L..R] exists in the given strings, otherwise output "No".

Sample Input
1
2
ab
ioc
6
Q 0 2
Q 3 4
C 1 o
C 4 b
Q 0 2
Q 3 4

Sample Output
Case #1:
No
No
Yes
Yes

/**
hdu 3973  线段树单点更新区间求值+字符串hash

*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <set>
using namespace std;

const int maxn=100010;
const int seed=31;
typedef unsigned long long LL;

struct note
{
int l,r;
LL hashes;
}tree[maxn*4];
char str[2000100];
LL Hash[maxn];
int n;

void init()
{
Hash[0]=1;
for(int i=1;i<maxn;i++)
{
Hash[i]=Hash[i-1]*seed;
}
}
LL get_hash()
{
int len=strlen(str);
LL sum=0;
for(int i=0;i<len;i++)
sum=sum*seed+str[i]-'a'+1;
return sum;
}

void build(int l,int r,int root)
{
tree[root].l=l;
tree[root].r=r;
if(l==r)
{
tree[root].hashes=str[l]-'a'+1;
return;
}
int mid=(l+r)/2;
build(l,mid,root<<1);
build(mid+1,r,root<<1|1);
tree[root].hashes=tree[root<<1].hashes*Hash[r-mid]+tree[root<<1|1].hashes;
}

void update(int l,int root)
{
if(tree[root].l==tree[root].r)
{
tree[root].hashes=str[l]-'a'+1;
return;
}
int mid=(tree[root].l+tree[root].r)>>1;
if(l<=mid) update(l,root<<1);
else  update(l,root<<1|1);
tree[root].hashes=tree[root<<1].hashes*Hash[tree[root].r-mid]+tree[root<<1|1].hashes;
}

LL query(int l,int r,int root)
{
//  printf("**\n");
if(tree[root].l==l&&tree[root].r==r)
return tree[root].hashes;
int mid=(tree[root].r+tree[root].l)>>1;
if(r<=mid)return query(l,r,root<<1);
else if(l>mid)return query(l,r,root<<1|1);
return query(l,mid,root<<1)*Hash[r-mid]+query(mid+1,r,root<<1|1);
}
int main()
{
int T,tt=0;
init();
scanf("%d",&T);
while(T--)
{
printf("Case #%d:\n",++tt);
scanf("%d",&n);
set<LL>mp;
for(int i=0;i<n;i++)
{
scanf("%s",str);
mp.insert(get_hash());
}
scanf("%s",str);
int len=strlen(str);
build(0,len-1,1);
int q;
scanf("%d",&q);
for(int i=1;i<=q;i++)
{
char c[5];
scanf("%s",c);
if(c[0]=='C')
{
int a;
char b[10];
scanf("%d%s",&a,b);
str[a]=b[0];
update(a,1);
}
else
{
int l,r;
scanf("%d%d",&l,&r);
if(mp.find(query(l,r,1))!=mp.end())
printf("Yes\n");
else
printf("No\n");
}
}
}
return 0;
}


1. 第一句可以忽略不计了吧。从第二句开始分析，说明这个花色下的所有牌都会在其它里面出现，那么还剩下♠️和♦️。第三句，可以排除2和7，因为在两种花色里有。现在是第四句，因为♠️还剩下多个，只有是♦️B才能知道答案。