南邮 OJ 1620 字符串
字符串
时间限制(普通/Java) :
20000 MS/ 3000 MS 运行内存限制 : 65536 KByte
总提交 : 53 测试通过 : 15
总提交 : 53 测试通过 : 15
比赛描述
已知有两个字串 A$, B$ 及一组字串变换的规则(至多6个规则):
A1$ -> B1$
A2$ -> B2$
规则的含义为：在 A$中的子串 A1$ 可以变换为 B1$、A2$ 可以变换为 B2$ …。
例如：A$＝'abcd' B$＝'xyz'
变换规则为：
'abc'->'xu' 'ud'->'y' 'y'->'yz'
则此时，A$ 可以经过一系列的变换变为 B$，其变换的过程为：
'abcd'->'xud'->'xy'->'xyz'
共进行了三次变换，使得 A$ 变换为B$。
输入
键盘输人文件名。文件格式如下：
A$ B$
A1$ B1$ \
A2$ B2$ |-> 变换规则
... ... /
所有字符串长度的上限为 20。
输出
输出至屏幕。格式如下：
若在 10 步(包含 10步)以内能将 A$ 变换为 B$,则输出最少的变换步数;否则输出"NO ANSWER!"
样例输入
abcd wyz
abc xu
ud y
y yz
样例输出
3
题目来源
NUPT
/* Time Limit Exceed at Test 4
#include<iostream>
#include<string>
using namespace std;
#define MAX_STEP 10
int step,len;
string A,B,a[6],b[6];
void dfs(string A,int cs){
if(cs>step || cs>MAX_STEP){
return;
}
int i,j,temp;
string s;
for(i=0;i<(int)A.length();i++){
temp = (int)A.size()-i;
for(j=0;j<len;j++){
if(temp>=(int)a[j].length()){
s = A.substr(i,a[j].size());
if(s==a[j]){
s = A.substr(0,i)+b[j]+A.substr(i+a[j].size(),A.size()-i-a[j].size());
if(s==B){
step = cs;
}else{
dfs(s,cs+1);
}
}
}
}
}
}
int main(){
// freopen("test.txt","r",stdin);
cin>>A>>B;
step = INT_MAX;
len = 0;
while(cin>>a[len]>>b[len]){
len++;
}
dfs(A,1);
if(step!=INT_MAX){
printf("%d\n",step);
}else{
printf("NO ANSWER!\n");
}
}
*/
/* 3MS
#include<iostream>
using namespace std;
struct node{
char s[30];
int dep;
} list1[5010],list2[5010];
char a[7][30],b[7][30];
int n;
bool check(char *s1,char *s2){
int len1=strlen(s1),len2=strlen(s2);
if (len1!=len2)
return false;
for (int i=0;i<strlen(s1);i++)
if (s1[i]!=s2[i])
return false;
return true;
}
bool pan1(char *s,int i,int x){
int len=strlen(a[x]);
for (int j=i;j<i+len;j++)
if (s[j]!=a[x][j-i])
return false;
return true;
}
bool pan2(char *s,int i,int x){
int len=strlen(b[x]);
for (int j=i;j<i+len;j++)
if (s[j]!=b[x][j-i]) return false;
return true;
}
void bfs(){
int head1,tail1,head2,tail2,i,j,k,l;
head1=tail1=head2=tail2=1;
while(head1<=tail1 && head2<=tail2){
if(list1[head1].dep+list2[head2].dep>10){
printf("NO ANSWER!\n");
return ;
}
int len1=strlen(list1[head1].s);
for( i=0;i<len1;i++){
for( j=1;j<=n;j++){
if(pan1(list1[head1].s,i,j)){
tail1++;
for(k=0;k<i;k++){
list1[tail1].s[k]=list1[head1].s[k];
}
int len2=strlen(b[j]);
for(l=0;l<len2;l++,k++){
list1[tail1].s[k]=b[j][l];
}
len2=strlen(list1[head1].s);
for(l=i+strlen(a[j]);l<=len2;l++,k++){
list1[tail1].s[k]=list1[head1].s[l];
}
list1[tail1].s[k]='\0';
list1[tail1].dep=list1[head1].dep+1;
for (k=1;k<=tail2;k++){
if (check(list1[tail1].s,list2[k].s)){
printf("%d\n",list1[tail1].dep+list2[k].dep);
return ;
}
}
}
}
}
len1=strlen(list2[head2].s);
for(i=0;i<len1;i++){
for(j=1;j<=n;j++){
if(pan2(list2[head2].s,i,j)){
tail2++;
for(k=0;k<i;k++){
list2[tail2].s[k]=list2[head2].s[k];
}
int len2=strlen(a[j]);
for(l=0;l<len2;l++,k++){
list2[tail2].s[k]=a[j][l];
}
len2=strlen(list2[head2].s);
for(l=i+strlen(b[j]);l<=len2;l++,k++){
list2[tail2].s[k]=list2[head2].s[l];
}
list2[tail2].s[k]='\0';
list2[tail2].dep=list2[head2].dep+1;
for (k=1;k<=tail1;k++){
if (check(list1[k].s,list2[tail2].s))
{
printf("%d\n",list1[k].dep+list2[tail2].dep);
return ;
}
}
}
}
}
head1++; head2++;
}
printf("NO ANSWER!\n");
}
int main(){
scanf("%s%s",list1[1].s,list2[1].s);
n=1;
while (scanf("%s%s",a[n],b[n])!=EOF) n++;
n--;
list1[1].dep=list2[1].dep=0;
bfs();
return 0;
}
*/
/* Wrong Answer at Test 4
#include<iostream>
#define MAX_LEN 30
#define MAX_DEP 10
#define MAX_N 5000
struct queueNode{
char c[MAX_LEN];
int dep;
}que[MAX_N];
int qHead,qTail;
char a[6][MAX_LEN];
char b[6][MAX_LEN];
char target[MAX_LEN];
int aSize;
//c2字符串等于c1前面同等长度的字符串
bool hasSub(char *c1,char *c2){
while(*c2 && *c1==*c2){
c1++;
c2++;
}
if(*c2){
return 0;
}
return 1;
}
int main(){
// freopen("test.txt","r",stdin);
char *p,*q;
int depth,len1,len2,i,j,k,l;
scanf("%s%s",que[qTail++].c,target);
if(strcmp(que[qHead].c,target)==0){
printf("0\n");
return 0;
}
while(scanf("%s%s",a[aSize],b[aSize])==2){
aSize++;
}
while(qHead<=qTail){
depth = que[qHead].dep+1;
if(depth>MAX_DEP){
break;
}
p = que[qHead++].c;
len1 = (int)strlen(p);
for(i=0;i<len1;i++){
for(j=0;j<aSize;j++){
if(hasSub(p+i,a[j])){
q = que[qTail].c;
que[qTail++].dep = depth;
for(k=0;k<i;k++){
q[k] = p[k]; //原字符串前面的部分
}
len2 = (int)strlen(b[j]);
for(l=0;l<len2;l++,k++){ //被替换掉的部分
q[k] = b[j][l];
}
len2 = (int)strlen(p);
for(l=i+(int)strlen(a[j]);l<len2;l++,k++){
q[k] = p[l];
}
q[k] = 0;
if(strcmp(q,target)==0){
printf("%d\n",depth);
return 0;
}
}
}
}
}
printf("NO ANSWER!\n");
}
*/
/*
// Runtime Error at Test 5
//(ACCESS_VIOLATION)
//由于数据范围的原因，单项搜索就不行,必须要双向搜索
#include<iostream>
using namespace std;
struct node{
char s[300];
int dep;
} list1[50100];
char a[7][30],b[7][30],target[30];
int n;
bool pan1(char *s,int i,int x){
int len=strlen(a[x]);
for (int j=i;j<i+len;j++)
if (s[j]!=a[x][j-i])
return false;
return true;
}
void bfs(){
int head1,tail1,i,j,k,l;
head1=tail1=1;
while(head1<=tail1){
if(list1[head1].dep>10){
printf("NO ANSWER!\n");
return ;
}
int len1=strlen(list1[head1].s);
for( i=0;i<len1;i++){
for( j=1;j<=n;j++){
if(pan1(list1[head1].s,i,j)){
tail1++;
for(k=0;k<i;k++){
list1[tail1].s[k]=list1[head1].s[k];
}
int len2=strlen(b[j]);
for(l=0;l<len2;l++,k++){
list1[tail1].s[k]=b[j][l];
}
len2=strlen(list1[head1].s);
for(l=i+strlen(a[j]);l<=len2;l++,k++){
list1[tail1].s[k]=list1[head1].s[l];
}
list1[tail1].s[k]='\0';
list1[tail1].dep=list1[head1].dep+1;
if (strcmp(list1[tail1].s,target)==0){
printf("%d\n",list1[tail1].dep);
return ;
}
}
}
}
head1++;
}
printf("NO ANSWER!\n");
}
int main(){
// freopen("test.txt","r",stdin);
scanf("%s%s",list1[1].s,target);
n=1;
while (scanf("%s%s",a[n],b[n])!=EOF) n++;
n--;
list1[1].dep=0;
bfs();
return 0;
}
*/
/* WA4
#include<iostream>
#define MAX_LEN 30
#define MAX_DEP 10
#define MAX_N 5000
struct queueNode{
char c[MAX_LEN];
int dep;
}que1[MAX_N],que2[MAX_N];
int head1,tail1,head2,tail2;
char a[6][MAX_LEN];
char b[6][MAX_LEN];
int aSize;
//c2字符串等于c1前面同等长度的字符串
bool hasSub(char *c1,char *c2){
while(*c2 && *c1==*c2){
c1++;
c2++;
}
if(*c2){
return 0;
}
return 1;
}
int main(){
freopen("test.txt","r",stdin);
char *p,*q;
int depth1,depth2,len1,len2,i,j,k,l;
scanf("%s%s",que1[tail1++].c,que2[tail2++].c);
if(strcmp(que1[head1].c,que2[head2].c)==0){
printf("0\n");
return 0;
}
while(scanf("%s%s",a[aSize],b[aSize])==2){
aSize++;
}
while(head1<=tail1 && head2<=tail2){
depth1 = que1[head1].dep+1;
depth2 = que2[head2].dep+1;
if(depth1+depth2 > MAX_DEP){
break;
}
if(tail1-head1 <= tail2-head2){
p = que1[head1++].c;
len1 = (int)strlen(p);
for(i=0;i<len1;i++){
for(j=0;j<aSize;j++){
if(hasSub(p+i,a[j])){
q = que1[tail1].c;
que1[tail1++].dep = depth1;
for(k=0;k<i;k++){
q[k] = p[k]; //原字符串前面的部分
}
len2 = (int)strlen(b[j]);
for(l=0;l<len2;l++,k++){ //被替换掉的部分
q[k] = b[j][l];
}
len2 = (int)strlen(p);
for(l=i+(int)strlen(a[j]);l<len2;l++,k++){
q[k] = p[l];
}
q[k] = 0;
for(k=head2;k<tail2;k++){
if(strcmp(q,que2[k].c)==0){
printf("%d\n",depth1+que2[k].dep);
return 0;
}
}
}
}
}
}
if(tail2-head2 <= tail1-head1){
p = que2[head2++].c;
len1 = (int)strlen(p);
for(i=0;i<len1;i++){
for(j=0;j<aSize;j++){
if(hasSub(p+i,b[j])){
q = que2[tail2].c;
que2[tail2++].dep = depth2;
for(k=0;k<i;k++){
q[k] = p[k]; //原字符串前面的部分
}
len2 = (int)strlen(a[j]);
for(l=0;l<len2;l++,k++){ //被替换掉的部分
q[k] = a[j][l];
}
len2 = (int)strlen(p);
for(l=i+(int)strlen(b[j]);l<len2;l++,k++){
q[k] = p[l];
}
q[k] = 0;
for(k=head1;k<tail1;k++){
if(strcmp(q,que1[k].c)==0){
printf("%d\n",depth2+que1[k].dep);
return 0;
}
}
}
}
}
}
}
printf("NO ANSWER!\n");
}
*/
/* AC 0MS
#include<iostream>
using namespace std;
struct node{
char s[30];
int dep;
} list1[5010],list2[5010];
char a[7][30],b[7][30];
int n;
bool check(char *s1,char *s2){
while(*s1 && *s2 && *s1==*s2){
s1++;
s2++;
}
if(*s1==0 && *s2==0){
return 1;
}
return 0;
}
bool pan1(char *s,int i,int x){
int len=strlen(a[x]);
for (int j=i;j<i+len;j++)
if (s[j]!=a[x][j-i])
return false;
return true;
}
bool pan2(char *s,int i,int x){
int len=strlen(b[x]);
for (int j=i;j<i+len;j++)
if (s[j]!=b[x][j-i]) return false;
return true;
}
void bfs(){
int head1,tail1,head2,tail2,i,j,k,l;
head1=tail1=head2=tail2=1;
while(head1<=tail1 && head2<=tail2){
if(list1[head1].dep+list2[head2].dep>10){
printf("NO ANSWER!\n");
return ;
}
int len1=strlen(list1[head1].s);
for( i=0;i<len1;i++){
for( j=1;j<=n;j++){
if(pan1(list1[head1].s,i,j)){
tail1++;
for(k=0;k<i;k++){
list1[tail1].s[k]=list1[head1].s[k];
}
int len2=strlen(b[j]);
for(l=0;l<len2;l++,k++){
list1[tail1].s[k]=b[j][l];
}
len2=strlen(list1[head1].s);
for(l=i+strlen(a[j]);l<=len2;l++,k++){
list1[tail1].s[k]=list1[head1].s[l];
}
list1[tail1].s[k]='\0';
list1[tail1].dep=list1[head1].dep+1;
for (k=1;k<=tail2;k++){
if (check(list1[tail1].s,list2[k].s)){
printf("%d\n",list1[tail1].dep+list2[k].dep);
return ;
}
}
}
}
}
len1=strlen(list2[head2].s);
for(i=0;i<len1;i++){
for(j=1;j<=n;j++){
if(pan2(list2[head2].s,i,j)){
tail2++;
for(k=0;k<i;k++){
list2[tail2].s[k]=list2[head2].s[k];
}
int len2=strlen(a[j]);
for(l=0;l<len2;l++,k++){
list2[tail2].s[k]=a[j][l];
}
len2=strlen(list2[head2].s);
for(l=i+strlen(b[j]);l<=len2;l++,k++){
list2[tail2].s[k]=list2[head2].s[l];
}
list2[tail2].s[k]='\0';
list2[tail2].dep=list2[head2].dep+1;
for (k=1;k<=tail1;k++){
if (check(list1[k].s,list2[tail2].s))
{
printf("%d\n",list1[k].dep+list2[tail2].dep);
return ;
}
}
}
}
}
head1++; head2++;
}
printf("NO ANSWER!\n");
}
int main(){
scanf("%s%s",list1[1].s,list2[1].s);
n=1;
while (scanf("%s%s",a[n],b[n])!=EOF) n++;
n--;
list1[1].dep=list2[1].dep=0;
bfs();
return 0;
}
*/
// AC 0MS
#include<iostream>
using namespace std;
struct node{
char s[30];
int dep;
} list1[5010],list2[5010];
char a[7][30],b[7][30];
int n;
//判断s1、s2字符串是否相等
bool equal(char *s1,char *s2){
while(*s1 && *s2 && *s1==*s2){
s1++;
s2++;
}
if(*s1==0 && *s2==0){
return 1;
}
return 0;
}
//判断s2是否为s1的前缀
bool preEqual(char *c1,char *c2){
while(*c2 && *c1==*c2){
c1++;
c2++;
}
if(*c2){
return 0;
}
return 1;
}
void bfs(){
int head1,tail1,head2,tail2,i,j,k,l;
head1=tail1=head2=tail2=1;
while(head1<=tail1 && head2<=tail2){
if(list1[head1].dep+list2[head2].dep>10){
printf("NO ANSWER!\n");
return ;
}
int len1=strlen(list1[head1].s);
for( i=0;i<len1;i++){
for( j=1;j<=n;j++){
if(preEqual(list1[head1].s+i,a[j])){
tail1++;
for(k=0;k<i;k++){
list1[tail1].s[k]=list1[head1].s[k];
}
int len2=strlen(b[j]);
for(l=0;l<len2;l++,k++){
list1[tail1].s[k]=b[j][l];
}
len2=strlen(list1[head1].s);
for(l=i+strlen(a[j]);l<=len2;l++,k++){
list1[tail1].s[k]=list1[head1].s[l];
}
list1[tail1].s[k]=0;
list1[tail1].dep=list1[head1].dep+1;
for (k=1;k<=tail2;k++){
if (equal(list1[tail1].s,list2[k].s)){
printf("%d\n",list1[tail1].dep+list2[k].dep);
return ;
}
}
}
}
}
len1=strlen(list2[head2].s);
for(i=0;i<len1;i++){
for(j=1;j<=n;j++){
if(preEqual(list2[head2].s+i,b[j])){
tail2++;
for(k=0;k<i;k++){
list2[tail2].s[k]=list2[head2].s[k];
}
int len2=strlen(a[j]);
for(l=0;l<len2;l++,k++){
list2[tail2].s[k]=a[j][l];
}
len2=strlen(list2[head2].s);
for(l=i+strlen(b[j]);l<=len2;l++,k++){
list2[tail2].s[k]=list2[head2].s[l];
}
list2[tail2].s[k]=0;
list2[tail2].dep=list2[head2].dep+1;
for (k=1;k<=tail1;k++){
if (equal(list1[k].s,list2[tail2].s)){
printf("%d\n",list1[k].dep+list2[tail2].dep);
return ;
}
}
}
}
}
head1++;
head2++;
}
printf("NO ANSWER!\n");
}
int main(){
// freopen("test.txt","r",stdin);
scanf("%s%s",list1[1].s,list2[1].s);
n=1;
while (scanf("%s%s",a[n],b[n])!=EOF){
n++;
}
n--;
list1[1].dep=list2[1].dep=0;
bfs();
return 0;
}