解題區/解題區 - UVa

UVa 1063 - Marble Game

contents

  1. 1. Problem
  2. 2. Sample Input
  3. 3. Sample Output
  4. 4. Solution

Problem

旋轉盤面，讓每顆球落入屬於自己的洞，請問至少需要幾次旋轉。

  • 球不能翻過牆壁、不能跨越其他球、不能飛過空洞。
  • 球不能離開盤面
  • 每一格最多只有一顆球
  • 當球落入空洞時，這個空洞相當於被填滿，其他的球可以經過這個被填滿的洞，但無法再次落入該洞。

Sample Input

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4 3 1
0 1 
1 0 
1 2 
2 3 
2 1 
3 2 
1 1 1 2
3 2 2
0 0 
0 1 
0 2 
2 0 
2 0 1 0
2 0 2 1
0 0 0

Sample Output

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Case 1: 5 moves 
Case 2: impossible

Solution

首先有可能落錯空洞，模擬時必須特別小心這種非法的旋轉。在一次操作中，有可能在落入空洞的瞬間，下一個球剛好可以通過這個填滿的洞。

將每個球的座標壓成狀態，進行 Bfs 搜索。

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#include <stdio.h>
#include <queue>
#include <map>
#include <string.h>
#include <algorithm>
using namespace std;
int N, W;
struct state {
    vector< pair<int, int> > xy;
    bool operator<(const state &a) const {
        for (int i = 0; i < xy.size(); i++)
            if (xy[i] != a.xy[i])
                return xy[i] < a.xy[i];
        return false;
    }
    int isComplete() {
        int ok = 1;
        for (int i = 0; i < xy.size() && ok; i++)
            ok &= xy[i].first < 0;
        return ok;
    } 
};
int g[4][4][4];
pair<int, int> goal[64];
const int dx[] = {0, 0, 1, -1};
const int dy[] = {1, -1, 0, 0};
int isValid(int x, int y) {
    return x >= 0 && y >= 0 && x < W && y < W;
}
state eraseGoal(state u) {
    for (int i = 0; i < u.xy.size(); i++) {
        if (u.xy[i] == goal[i])
            u.xy[i] = pair<int, int>(-1, -1);
    }
    return u;
}
state rotateMap(state u, int dir, int& ok) {
    ok = 1;
    int update = 1, tx, ty, x, y;
    int used[4][4] = {}, usedg[4][4] = {};
    for (int i = 0; i < u.xy.size(); i++) {
        x = u.xy[i].first, y = u.xy[i].second;
        if (x == -1)	continue;
        used[x][y] = 1;
        x = goal[i].first, y = goal[i].second;
        usedg[x][y] = i+1;
    }
    do {
        update = 0;
        for (int i = 0; i < u.xy.size(); i++) {
            while (u.xy[i].first >= 0) {
                x = u.xy[i].first, y = u.xy[i].second;
                tx = x + dx[dir], ty = y + dy[dir];
                if (isValid(tx, ty) && !g[x][y][dir] && usedg[tx][ty] && usedg[tx][ty] != i+1)
                    ok = 0;
                if (isValid(tx, ty) && !g[x][y][dir] && (!used[tx][ty] || goal[i] == make_pair(tx, ty))) {
                    used[x][y] = 0, used[tx][ty] = 1;
                    u.xy[i] = pair<int, int>(tx, ty);
                    update = 1;
                    if (goal[i] == pair<int, int>(tx, ty)) {
                        u.xy[i] = pair<int, int>(-1, -1);
                        used[tx][ty] = 0, usedg[tx][ty] = 0;
                    }
                } else {
                    break;
                }
            }
        } 
    } while (update);
    return u;
}
void print(state u) {
    int used[4][4] = {}, x, y;
    for (int i = 0; i < u.xy.size(); i++) {
        x = u.xy[i].first, y = u.xy[i].second;
        if (x == -1)
            continue;
        used[x][y] = i+1;
    }
    for (int i = 0; i < W; i++) {
        for (int j = 0; j < W; j++)
            printf("%d ", used[i][j]);
        puts("");
    }
    puts("--");
}
int bfs(state init) {
    state u, v;
    queue<state> Q;
    map<state, int> R;
    int f;
    
    init = eraseGoal(init);
    Q.push(init), R[init] = 0;
//	print(init);
    if (init.isComplete())
        return 0;
    
    while (!Q.empty()) {
        u = Q.front(), Q.pop();
        int step = R[u];
        
//		print(u);
//		printf("step %d\n", step);
        for (int i = 0; i < 4; i++) {
            v = rotateMap(u, i, f);
            v = eraseGoal(v);
            if (!f || R.count(v))		continue;
            if (v.isComplete())
                return step + 1;
            R[v] = step + 1;
//			print(v);
            Q.push(v);
        }
//		puts("--------------");
//		getchar();
    }
    return -1;
}
int main() {
    int x, y, tx, ty, M;
    int sx, sy, ex, ey;
    int cases = 0;
    while (scanf("%d %d %d", &W, &N, &M) == 3 && N+W+M) {
        state init;
        memset(g, 0, sizeof(g));
        
        for (int i = 0; i < N; i++) {
            scanf("%d %d", &x, &y);
            init.xy.push_back(pair<int, int>(x, y));
        }
        for (int i = 0; i < N; i++) {
            scanf("%d %d", &x, &y);
            goal[i] = pair<int, int>(x, y);
        }
        for (int i = 0; i < M; i++) {
            scanf("%d %d %d %d", &sx, &sy, &ex, &ey);
            for (int j = 0; j < 4; j++) {
                tx = sx + dx[j], ty = sy + dy[j];
                if (tx == ex && ty == ey)
                    g[sx][sy][j] = 1;
                tx = ex + dx[j], ty = ey + dy[j];
                if (tx == sx && ty == sy)
                    g[ex][ey][j] = 1;
            }
        }
        
        int step = bfs(init);
        
        printf("Case %d: ", ++cases);
        if (step < 0)
            puts("impossible");
        else
            printf("%d moves\n", step);
        puts("");
    }
    return 0;
}
/*
3 1 5
1 2
1 0
2 1 2 2
0 0 1 0
0 1 1 1
0 2 1 2
1 1 2 1
4 3 1
0 1 
1 0 
1 2 
2 3 
2 1 
3 2 
1 1 1 2
3 2 2
0 0 
0 1 
0 2 
2 0 
2 0 1 0
2 0 2 1
*/