题意
中文题意
思路
做这题的前置技能学习
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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。 -
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
", ans);
else puts("No Solution!");
}
return 0;
}