题目描述
解决思路
- 启发函数：只需要搜索非常少的状态，就可以搜到从起点到终点的最短路径
- 估价函数：当前状态中每个数与它的目标位置的曼哈顿距离之和
- A*算法
优先级为：从起点到当前点的真实距离 + 从当前点到终点的估计距离
题目代码
#include
#include
#include
#include
#include
using namespace std;
typedef pair PIS;
int f(string state) // 估价函数
{
int res = 0;
for(int i = 0; i < state.size(); i ++ )
if(state[i] != 'x')
{
int t = state[i] - '1';
res += abs(i / 3 - t / 3) + abs(i % 3 - t % 3);
}
return res;
}
string bfs(string start)
{
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
char op[4] = {'u', 'r', 'd', 'l'};
string end = "12345678x";
unordered_map dist;
unordered_map> prev;
priority_queue, greater> heap;
heap.push({f(start), start});
dist[start] = 0;
while(heap.size())
{
auto t = heap.top();
heap.pop();
string state = t.second;
if(state == end) break;
int step = dist[state];
int x, y;
for(int i = 0; i < state.size(); i ++ )
if(state[i] == 'x')
{
x = i / 3, y = i % 3;
break;
}
string source = state;
for(int i = 0; i < 4; i ++ )
{
int a = x + dx[i], b = y + dy[i];
if(a >= 0 && a < 3 && b >= 0 && b < 3)
{
swap(state[x * 3 + y], state[a * 3 + b]);
if(!dist.count(state) || dist[state] > step + 1)
{
dist[state] = step + 1;
prev[state] = {source, op[i]};
heap.push({dist[state] + f(state), state});
}
swap(state[x * 3 + y], state[a * 3 + b]);
}
}
}
string res;
while (end != start)
{
res += prev[end].second;
end = prev[end].first;
}
reverse(res.begin(), res.end());
return res;
}
int main()
{
string g, c, seq;
while(cin >> c)
{
g += c;
if(c != "x") seq += c;
}
int t = 0;
for(int i = 0; i < seq.size(); i ++ )
for(int j = i + 1; j < seq.size(); j ++ )
if(seq[i] > seq[j])
t ++ ;
if(t % 2) puts("unsolvable");
else cout << bfs(g) << endl;
return 0;
}
Original: https://www.cnblogs.com/esico/p/16483908.html
Author: esico
Title: AcWing 179. 八数码（搜索）
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