hihoCoder 1312：搜索三·启发式搜索（A* + 康托展开）

题目链接

题意

中文题意

思路

做这题的前置技能学习

1. 康托展开
   
   这个东西我认为就是在排列组合问题上的Hash算法，可以压缩空间。

2. A*搜索。
   
   这里我使用了像k短路一样的做法，从最终状态倒回去预处理一遍距离，但是跑了0.8s，可能是预处理花费的时间太多了。有些人用曼哈顿距离估价，跑了0.2s。

#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 1e5 + 10;
typedef long long LL;
struct Node {
    int x, y, f, g, kt, mm[4][4];
    bool operator < (const Node &rhs) const {
        return f > rhs.f;
    }
};
int init[4][4] = {
    { 1, 2, 3 },
    { 4, 5, 6 },
    { 7, 8, 0 }
};
int mp[4][4], ans, h[500001], vis[500001], target;
int f[10], dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};

int kangtuo(int mp[4][4]) {
    int sum = 0;
    for(int i = 0; i < 9; i++) {
        int now = 0;
        for(int j = i + 1; j < 9; j++) {
            if(mp[i/3][i%3] > mp[j/3][j%3]) now++;
        }
        sum += now * f[9-i-1];
    } return sum + 1;
}

void BFS(int x, int y) {
    memset(h, INF, sizeof(h));
    queue<Node> que;
    Node now = (Node) {x, y, 0, 0};
    for(int i = 0; i < 3; i++) for(int j = 0; j < 3; j++)
        now.mm[i][j] = init[i][j];
    h[target] = 0;
    que.push(now);
    while(!que.empty()) {
        Node now = que.front(); que.pop();
        int x = now.x, y = now.y, f = now.f;
        for(int i = 0; i < 4; i++) {
            now.x = x + dx[i], now.y = y + dy[i];
            if(now.x < 0 || now.x > 2 || now.y < 0 || now.y > 2) continue;
            swap(now.mm[x][y], now.mm[now.x][now.y]);
            int nf = f + 1, nkt = kangtuo(now.mm);
            now.f = nf;
            if(h[nkt] == INF) {
                h[nkt] = nf;
                que.push(now);
            }
            swap(now.mm[x][y], now.mm[now.x][now.y]);
        }
    }
}

void YCL() {
    BFS(2, 2);
}

int Astar(int x, int y) {
    priority_queue<Node> que;
    memset(vis, 0, sizeof(vis));
    vis[kangtuo(mp)] = 1;
    Node now = (Node) { x, y, 0, 0, kangtuo(mp)};
    for(int i = 0; i < 3; i++)
        for(int j = 0; j < 3; j++)
            now.mm[i][j] = mp[i][j];
    que.push(now);
    while(!que.empty()) {
        now = que.top(); que.pop();
        int x = now.x, y = now.y, f = now.f, g = now.g, kt = now.kt;
        if(kt == target) return f;
        for(int i = 0; i < 4; i++) {
            int nx = x + dx[i], ny = y + dy[i];
            if(nx < 0 || nx > 2 || ny < 0 || ny > 2) continue;
            swap(now.mm[x][y], now.mm[nx][ny]);
            int nkt = kangtuo(now.mm);
            now.x = nx, now.y = ny, now.g = g + 1, now.f = now.g + h[nkt], now.kt = nkt;
            if(!vis[nkt]) {
                vis[nkt] = 1;
                que.push(now);
            }
            swap(now.mm[x][y], now.mm[nx][ny]);
        }
    } return -1;
}

int main() {
    f[0] = 1;
    for(int i = 1; i < 9; i++) f[i] = f[i-1] * (i);
    target = kangtuo(init);
    YCL();
    int t; scanf("%d", &t);
    while(t--) {
        int ix, iy;
        for(int i = 0; i < 3; i++)
            for(int j = 0; j < 3; j++) {
                scanf("%d", &mp[i][j]);
                if(mp[i][j] == 0) ix = i, iy = j;
            }
        ans = Astar(ix, iy);
        if(~ans) printf("%d\n", ans);
        else puts("No Solution!");
    }
    return 0;
}