題目鏈接
題意
中文題意
思路
做這題的前置技能學(xué)習(xí)
康托展開
這個東西我認(rèn)為就是在排列組合問題上的Hash算法，可以壓縮空間。
A*搜索。
這里我使用了像k短路一樣的做法，從最終狀態(tài)倒回去預(yù)處理一遍距離，但是跑了0.8s，可能是預(yù)處理花費的時間太多了。有些人用曼哈頓距離估價，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;
}
轉(zhuǎn)載于:https://www.cnblogs.com/fightfordream/p/7590517.html
總結(jié)
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