标签：codeforces
Professor GukiZ doesn‘t accept string as they are. He likes to swap some letters in string to obtain a new one.
GukiZ has strings a, b, and c. He wants to obtain string k by swapping some letters in a, so that k should contain as many non-overlapping substrings equal either to b or c as possible. Substring of string x is a string formed by consecutive segment of characters from x. Two substrings of string x overlap if there is position i in string x occupied by both of them.
GukiZ was disappointed because none of his students managed to solve the problem. Can you help them and find one of possible stringsk?
The first line contains string a, the second line contains string b, and the third line contains string c (1?≤?|a|,?|b|,?|c|?≤?105, where |s|denotes the length of string s).
All three strings consist only of lowercase English letters.
It is possible that b and c coincide.
Find one of possible strings k, as described in the problem statement. If there are multiple possible answers, print any of them.
aaa a b
aaa
pozdravstaklenidodiri niste dobri
nisteaadddiiklooprrvz
abbbaaccca ab aca
ababacabcc
In the third sample, this optimal solutions has three non-overlaping substrings equal to either b or c on positions 1?–?2 (ab), 3?–?4 (ab),5?–?7 (aca). In this sample, there exist many other optimal solutions, one of them would be acaababbcc.
题意：
给出一个3个串s1,s2,s3要求改变s1串字母的位置,使得s2,s3在在改变后的串出现的总次数尽量多，而且不算重叠部分
思路：
先枚举其中一串出现的次数，然后计算不同情况下另外一串能出现的最大次数，最后看总次数最大的情况
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <list>
#include <algorithm>
#include <climits>
using namespace std;
#define lson 2*i
#define rson 2*i+1
#define LS l,mid,lson
#define RS mid+1,r,rson
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 100005
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define lowbit(x) (x&-x)
const int mod = 1e9+7;
char s1[N],s2[N],s3[N];
int len1,len2,len3;
int h1[30],h2[30],h3[30];
int cnt2,cnt3,cnt;
int main()
{
int i,j,k;
while(~scanf("%s%s%s",s1,s2,s3))
{
len1 = strlen(s1);
len2 = strlen(s2);
len3 = strlen(s3);
MEM(h1,0);
MEM(h2,0);
MEM(h3,0);
for(i = 0; i<len1; i++)
h1[s1[i]-'a']++;
for(i = 0; i<len2; i++)
h2[s2[i]-'a']++;
for(i = 0; i<len3; i++)
h3[s3[i]-'a']++;
cnt = 0;
for(i=1;; i++)
{
int p = N;
for(j = 0; j<26; j++)
{
if(h1[j]-h2[j]*i<0)
{
p=0;
break;
}
}
if(!p) break;
int p2 = N;
for(j = 0; j<26; j++)
{
if(h3[j])
{
p2 = min(p2,(h1[j]-h2[j]*i)/h3[j]);
}
}
if(i+p2>cnt)
{
cnt = i+p2;
cnt2 = i;
cnt3 = p2;
}
}
for(i=1;; i++)
{
int p = N;
for(j = 0; j<26; j++)
{
if(h1[j]-h3[j]*i<0)
{
p=0;
break;
}
}
if(!p) break;
int p2 = N;
for(j = 0; j<26; j++)
{
if(h2[j])
{
p2 = min(p2,(h1[j]-h3[j]*i)/h2[j]);
}
}
if(i+p2>cnt)
{
cnt = i+p2;
cnt3 = i;
cnt2 = p2;
}
}
for(i = 0; i<cnt2; i++)
printf("%s",s2);
for(i = 0; i<cnt3; i++)
printf("%s",s3);
for(i = 0; i<26; i++)
{
for(j = 0; j<h1[i]-cnt2*h2[i]-cnt3*h3[i]; j++)
printf("%c",i+'a');
}
printf("\n");
}
return 0;
}
标签：codeforces
原文地址：http://blog.csdn.net/libin56842/article/details/46533913
