There are restrictions wide search, re-sentenced to an array (or anything else) must, be, is K-dimensional ah ah ah ......
that is, the state is determined by how many arguments, sentenced to heavy-dimensional array will be much of. Including pre-processing time, the array should be looking for is how many dimensions.
This seems to have written a blog in the back ...... think back about silly repair before.
Example:? What a ridiculous election
described
... In country Light Tower, a presidential election is going on There aretwo candidates, Mr. X1andMr X2, and both of them are not like good persons One is called aliar and the other is called a maniac. They tear (Chinese English word , meansdefame) each other on TV face to face, on newspaper, on internet ... on allkinds of media. The country is tore into two parts because the people whosupport X1are almost as many as the people who support X2.
After the election day, X1and X2get almost thesame number of votes. No one gets enough votes to win. According to the law ofthe country, the Great Judge must decide who will be the president. But thejudge doesn’t want to offend half population of the country, so he randomlychooses a 6 years old kid Tom and authorize him to pick the president. Soundsweird? But the democracy in Light Tower is just like that.
The poor or lucky little kid Tom doesn’t understand what is happening tohis country. But he has his way to do his job. Tom’s ao shu(Chinese Englishword, means some kind of weird math for kids) teacher just left him a puzzle afew days ago, Tom decide that he who solve that puzzle in a better way will bepresident. The ao shu teacher’s puzzle is like this:
Given a string which consists of five digits(‘0’-‘9’), like"02943", you should change “12345” into it by as few aspossible operations. There are 3 kinds of operations:
-
Swap two adjacent digits.
-
Increase a digit by one. If the result exceed 9, change it to it modulo10.
-
Double a digit. If the result exceed 9, change it to it modulo 10.
You can use operation 2 at most three times, and use operation 3 at mosttwice.
As a melon eater(Chinese English again, means bystander), whichcandidate do you support? Please help him solve the puzzle.
输入
There are no more than 100,000 test cases.
Each test case is a string which consists of 5 digits.
输出
For each case,print the minimum number of operations must be used to change “12345” into the given string. If there is no solution, print -1.
样例输入
12435
99999
12374
样例输出
1
-1
3
Error code ah I'm not afraid of shame warning:
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <map>
using namespace std;
int visited[100003];
struct state {
int num;
int op2;
int op3;
int opnum;
state () {}
state (int _num, int _op2, int _op3, int _opnum): num(_num), op2(_op2), op3(_op3), opnum(_opnum) {}
};
int op1(int num, int pos) {
char cur[10];
sprintf(cur, "%05d", num);
char tmp = cur[pos];
cur[pos] = cur[pos + 1];
cur[pos + 1] = tmp;
int ret = atoi(cur);
return ret;
}
int op2(int num, int pos) {
char cur[10];
sprintf(cur, "%05d", num);
int tmp = cur[pos] - '0';
tmp = (tmp + 1) % 10;
cur[pos] = char(tmp + '0');
int ret = atoi(cur);
return ret;
}
int op3(int num, int pos) {
char cur[10];
sprintf(cur, "%05d", num);
int tmp = cur[pos] - '0';
tmp = (tmp * 2) % 10;
cur[pos] = char(tmp + '0');
int ret = atoi(cur);
return ret;
}
int main() {
queue <state> Q;
map <int, int> S;
char input[10];
Q.push(state(12345, 0, 0, 0));
visited[12345] = 1;
S.insert(make_pair(12345, 0));
state cur;
while (!Q.empty()) {
cur = Q.front();
Q.pop();
if (cur.op3 < 2) {
for (int i = 0; i < 5; ++i) {
int newnum = op3(cur.num, i);
if (visited[newnum]) continue;
visited[newnum] = 1;
S.insert(make_pair(newnum, cur.opnum + 1));
Q.push(state(newnum, cur.op2, cur.op3 + 1, cur.opnum + 1));
}
}
if (cur.op2 < 3) {
for (int i = 0; i < 5; ++i) {
int newnum = op2(cur.num, i);
if (visited[newnum]) continue;
visited[newnum] = 1;
S.insert(make_pair(newnum, cur.opnum + 1));
Q.push(state(newnum, cur.op2 + 1, cur.op3, cur.opnum + 1));
}
}
for (int i = 0; i < 4; ++i) {
int newnum = op1(cur.num, i);
if (visited[newnum]) continue;
visited[newnum] = 1;
S.insert(make_pair(newnum, cur.opnum + 1));
Q.push(state(newnum, cur.op2, cur.op3, cur.opnum + 1));
}
}
while (cin >> input) {
typename map <int, int>::iterator it = S.find(atoi(input));
if (it == S.end()) cout << -1 << endl;
else cout << it->second << endl;
}
return 0;
}
AC Code (copied answers (Behind his) or else simply can not find ......)
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <map>
#include <algorithm>
using namespace std;
const int INF = 100003;
int visited[100003][4][3];
struct state {
int num;
int op2;
int op3;
int opnum;
state () {}
state (int _num, int _op2, int _op3, int _opnum): num(_num), op2(_op2), op3(_op3), opnum(_opnum) {}
};
int op1(int num, int pos) {
char cur[10];
sprintf(cur, "%05d", num);
char tmp = cur[pos];
cur[pos] = cur[pos + 1];
cur[pos + 1] = tmp;
int ret = atoi(cur);
return ret;
}
int op2(int num, int pos) {
char cur[10];
sprintf(cur, "%05d", num);
int tmp = cur[pos] - '0';
tmp = (tmp + 1) % 10;
cur[pos] = char(tmp + '0');
int ret = atoi(cur);
return ret;
}
int op3(int num, int pos) {
char cur[10];
sprintf(cur, "%05d", num);
int tmp = cur[pos] - '0';
tmp = (tmp * 2) % 10;
cur[pos] = char(tmp + '0');
int ret = atoi(cur);
return ret;
}
int main() {
for (int i = 0; i < INF; ++i) {
for (int j = 0; j < 4; ++j) {
for (int k = 0; k < 3; ++k) {
visited[i][j][k] = INF;
}
}
}
queue <state> Q;
map <int, int> S;
char input[10];
Q.push(state(12345, 0, 0, 0));
visited[12345][0][0] = 0;
S.insert(make_pair(12345, 0));
state cur;
while (!Q.empty()) {
cur = Q.front();
Q.pop();
if (cur.op3 < 2) {
for (int i = 0; i < 5; ++i) {
int newnum = op3(cur.num, i);
if (visited[newnum][cur.op2][cur.op3 + 1] != INF) continue;
visited[newnum][cur.op2][cur.op3 + 1] = cur.opnum + 1;
//S.insert(make_pair(newnum, cur.opnum + 1));
Q.push(state(newnum, cur.op2, cur.op3 + 1, cur.opnum + 1));
}
}
if (cur.op2 < 3) {
for (int i = 0; i < 5; ++i) {
int newnum = op2(cur.num, i);
if (visited[newnum][cur.op2 + 1][cur.op3] != INF) continue;
visited[newnum][cur.op2 + 1][cur.op3] = cur.opnum + 1;
//S.insert(make_pair(newnum, cur.opnum + 1));
Q.push(state(newnum, cur.op2 + 1, cur.op3, cur.opnum + 1));
}
}
for (int i = 0; i < 4; ++i) {
int newnum = op1(cur.num, i);
if (visited[newnum][cur.op2][cur.op3] != INF) continue;
visited[newnum][cur.op2][cur.op3] = cur.opnum + 1;
//S.insert(make_pair(newnum, cur.opnum + 1));
Q.push(state(newnum, cur.op2, cur.op3, cur.opnum + 1));
}
}
while (cin >> input) {
int ans = INF;
for (int i = 0; i < 4; ++i) {
for (int j = 0; j < 3; ++j) {
ans = min(ans, visited[atoi(input)][i][j]);
}
}
if (ans == INF) cout << -1 << endl;
else cout << ans << endl;
}
return 0;
}