UVA1533 Moving Pegs（bfs+hash）

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题意：

一个跳棋，给定一个空位。求最少步数且字典序最小使得跳到最后棋盘只剩一个旗子，且在初始的空位上。

思路：

BFS + hash判重，这题的思路还是比较好想的，但是我看网络上面的代码，都是打表+bfs，我直接把状态转换为2维数组，对6个方向进行bfs。

AC代码

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <vector>using namespace std;const int dx[] = {-1,-1, 0, 0, 1, 1};const int dy[] = {-1, 0,-1, 1, 0, 1};const int N = 1<<16;struct State {    int grid[5][5], num; //the last pos and the step    vector< pair<int,int> > path; //first from second to    //init    State() {        memset(grid, false, sizeof(grid));        num = 0;        path.clear();    }    //create function    State(int _grid[][5], int _num) {        memcpy(grid, _grid, sizeof(grid));        num = _num;    }    //hash    int Hash() {        int sum = 0, cnt = 1;        for(int i = 0; i < 5; i++)            for(int j = 0; j <= i; j++)                sum |= grid[i][j] << cnt++;        return sum;    }};vector< pair<int,int> > ans;int state[5][5], vis[N];int empty;void init() {    memset(vis, false, sizeof(vis));    memset(state, -1, sizeof(state));    int cnt = 1;    for(int i = 0; i < 5; i++) {        for(int j = 0; j <= i; j++) {            if(empty != cnt++) state[i][j] = 1;            else state[i][j] = 0;        }    }}int trans(int x, int y) {    return x*(x+1)/2 + (y+1);}bool check(int grid[][5]) {    int pos = 0;    for(int i = 0; i < 5; i++) {        for(int j = 0; j <= i; j++) {            if(grid[i][j]) pos = trans(i, j);        }    }    return pos == empty;}bool overBound(int x, int y) {    if(x < 0 || x >= 5 || y < 0 || y >= 5)        return true;    return false;}bool bfs() {    //init    queue<State> que;    que.push(State(state, 14));    State front = que.front(), tmpState;    vis[front.Hash()] = true;    int nx, ny, len;    while(!que.empty()) {        front = que.front(); que.pop();        if(front.num == 1 && check(front.grid)) {            ans = front.path; return true;        }        for(int x = 0; x < 5; x++) {            for(int y = 0; y <= x; y++) { //num                if(front.grid[x][y] == 0) continue;                for(int i = 0; i < 6; i++) { //direction                    nx = x + dx[i], ny = y + dy[i];                    if(!overBound(nx, ny)) {                        len = 1;                        tmpState = front;                        while(tmpState.grid[nx][ny] == 1 && !overBound(nx, ny)) {                            tmpState.grid[nx][ny] = 0;                            len++;                            nx += dx[i], ny += dy[i];                        }                    }                    if(tmpState.grid[nx][ny] == -1 || overBound(nx,ny) || len < 2) continue;                    tmpState.grid[nx][ny] = 1; tmpState.grid[x][y] = 0;                    tmpState.num -= (len-1);                    tmpState.path.push_back(make_pair(trans(x,y), trans(nx,ny)));                    int tmp = tmpState.Hash();                    if(!vis[tmp]) {                        vis[tmp] = true;                        que.push(tmpState);                    }                }            }        }    }    return false;}int main() {//  freopen("in.txt", "r", stdin);    int T;    scanf("%d", &T);    while(T--) {        scanf("%d", &empty);        init();        if(!bfs()) {            printf("IMPOSSIBLE\n");        }else {            printf("%d\n", ans.size());            printf("%d %d", ans[0].first, ans[0].second);            for(int i = 1; i < ans.size(); i++)                printf(" %d %d", ans[i].first, ans[i].second);            puts("");        }    }    return 0;}

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