Codeforces Round #338 (Div. 2) C. Running Track 字符串处理

题意：给两个字符串s1,s3。s3能否由s1或者s1的反字符串s2的某些连续的部分构成。例：s1 = “abc” s2 = “cbaabc” “cbaabc” = “cba” + “abc”.

思路：暴力就好，首先看”c”能不能在s1或者s2中找到，然后看”cb”能不能在s1或者s2中找到，如果能找到，继续这个过程。如果不能找到,”c”就是一个结果，”b”就是下一个字符串的第一个字符。

http://codeforces.com/contest/615/problem/C

/*********************************************    Problem : Codeforces    Author  : NMfloat    InkTime (c) NM . All Rights Reserved .********************************************/#include <map>#include <set>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>#define rep(i,a,b)  for(int i = (a) ; i <= (b) ; i ++) //遍历#define rrep(i,a,b) for(int i = (b) ; i >= (a) ; i --) //反向遍历#define repS(it,p) for(auto it = p.begin() ; it != p.end() ; it ++) //遍历一个STL容器#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next) //遍历u所连接的点#define cls(a,x)   memset(a,x,sizeof(a))#define eps 1e-8using namespace std;const int MOD = 1e9+7;const int INF = 0x3f3f3f3f;const int MAXN = 1e5+5;const int MAXE = 2e5+5;typedef long long LL;typedef unsigned long long ULL;int T,n,m,k;int fx[] = {0,1,-1,0,0};int fy[] = {0,0,0,-1,1};char s1[2105],s2[2105],s3[2105];char tmp[2105];int lens ;void get_s2() {    rep(i,0,lens-1) {        s2[lens-i-1] = s1[i];    }    s2[lens] = 0;}void input() {}int pos1 = 0 , pos2 = 0;int prepos1 , prepos2;int par[2105][2];int pos = 0;void get_pos() {    prepos1 = pos1 ;    prepos2 = pos2 ;    if(pos > lens) {        pos1 = pos2 = -1;        return ;    }    pos1 = strstr(s1,tmp) - s1;    pos2 = strstr(s2,tmp) - s2;}  void solve() {    lens = strlen(s1);    get_s2();    int len = strlen(s3);    int tot = 0;    pos1 = -1 ; pos2 = -1;    pos = 0;    int ok = 0;    s3[len] = '#';    rep(i,0,len) {        tmp[pos++] = s3[i];        tmp[pos] = 0;        get_pos();        if(pos1 >= 0 || pos2 >= 0) {        }        else {            ++ tot;            if(prepos1 < 0 && prepos2 <0) {                ok = 1;                break;            }            if(prepos1 >= 0) {                par[tot][0] = prepos1 + 1;                par[tot][1] = prepos1 + strlen(tmp) - 1;            }            else {                par[tot][0] = lens - prepos2 ;                par[tot][1] = lens - prepos2 - strlen(tmp) + 2;            }            pos = 0;            tmp[pos++] = s3[i];            tmp[pos] = 0;            get_pos();        }    }    if(!ok) {        printf("%d\n",tot);        rep(i,1,tot) {            printf("%d %d\n",par[i][0],par[i][1]);        }    }    else {        puts("-1");    }}int main(void) {    //freopen("a.in","r",stdin);    while(~scanf("%s%s",s1,s3)) {        input();        solve();    }    return 0;}