Codeforces Round #338 (Div. 2) C. Running Track

C. Running Track
time limit per test1 second
memory limit per test512 megabytes
inputstandard input
outputstandard output
A boy named Ayrat lives on planet AMI-1511. Each inhabitant of this planet has a talent. Specifically, Ayrat loves running, moreover, just running is not enough for him. He is dreaming of making running a real art.

First, he wants to construct the running track with coating t. On planet AMI-1511 the coating of the track is the sequence of colored blocks, where each block is denoted as the small English letter. Therefore, every coating can be treated as a string.

Unfortunately, blocks aren’t freely sold to non-business customers, but Ayrat found an infinite number of coatings s. Also, he has scissors and glue. Ayrat is going to buy some coatings s, then cut out from each of them exactly one continuous piece (substring) and glue it to the end of his track coating. Moreover, he may choose to flip this block before glueing it. Ayrat want’s to know the minimum number of coating s he needs to buy in order to get the coating t for his running track. Of course, he also want’s to know some way to achieve the answer.

Input
First line of the input contains the string s — the coating that is present in the shop. Second line contains the string t — the coating Ayrat wants to obtain. Both strings are non-empty, consist of only small English letters and their length doesn’t exceed 2100.

Output
The first line should contain the minimum needed number of coatings n or -1 if it’s impossible to create the desired coating.

If the answer is not -1, then the following n lines should contain two integers xi and yi — numbers of ending blocks in the corresponding piece. If xi ≤ yi then this piece is used in the regular order, and if xi > yi piece is used in the reversed order. Print the pieces in the order they should be glued to get the string t.

Sample test(s)
input
abc
cbaabc
output
2
3 1
1 3
input
aaabrytaaa
ayrat
output
3
1 1
6 5
8 7
input
ami
no
output
-1
Note
In the first sample string “cbaabc” = “cba” + “abc”.

In the second sample: “ayrat” = “a” + “yr” + “at”.

题目链接：http://codeforces.com/contest/615/problem/C

题目大意：
输入字符串s和t，t要由s的一些片段组合成，s的片段可颠倒过来，求出最少的片段数以及每个片段的起始和终结字母的编号。

解题思路：暴力模拟。注意不能用string，否则会t，可是增加一个字符串表示反过来的s串。

代码如下：

#include <cstdio>    #include <iostream>    #include <algorithm>    #include <cstring>    #include <string>    using namespace std;    int const maxn=2110;    char a[maxn],b[maxn],c[maxn];    int x[maxn],y[maxn];    int main(void)    {        scanf("%s",a);        scanf("%s",c);        int n=strlen(a);        for(int i=0;i<n;i++)            b[i]=a[n-i-1];        int m=strlen(c);        int ans=0;        for(int i=0;i<m;i++)        {            int ma=0,xx,yy;            for(int j=0;j<n;j++)            {                if(c[i]==a[j])                {                    int k=i;                    int h=j;                    while(c[k]==a[h])                    {                        k++;                        h++;                        if(k==m || h==n)                            break;                    }                    int l=k-i;                    if(ma<l)                    {                        ma=l;                        xx=j+1;                        yy=j+l;                             }                }               }            for(int j=0;j<n;j++)            {                if(c[i]==b[j])                {                    int k=i;                    int h=j;                    while(c[k]==b[h])                    {                        k++;                        h++;                        if(k==m || h==n)                            break;                    }                    int l=k-i;                    if(ma<l)                    {                        ma=l;                        xx=n-j;                        yy=n-(j+l-1);                               }                }               }            if(ma)            {                x[ans]=xx;                y[ans]=yy;                ans++;                i+=(ma-1);            }            else            {                ans=0;                break;            }        }        if(ans==0)        {            printf("-1\n");            return 0;        }        printf("%d\n",ans );        for(int i=0;i<ans;i++)            printf("%d %d\n",x[i],y[i] );    }