sicily 1133

最新推荐文章于 2016-10-18 23:27:26 发布

最新推荐文章于 2016-10-18 23:27:26 发布 · 52 阅读

本文介绍了一种使用广度优先搜索(BFS)解决九宫拼图问题的方法。通过预计算所有可能状态的最短步数，实现快速求解任意初始状态到目标状态的最少移动次数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

刚开始做是一组数据bfs一次，TLE，后来改成先全搜一次。。然后直接出答案

#include <iostream> #include <cstring> using namespace std; struct node { int s[9]; int step; }q[511777]; int ans[511777]; int init[9]; int tar[9] = {1, 2, 3, 4, 5, 6, 7, 8, 9}; int fac[9]; bool isUsed[511777]; void setFac() { fac[0] = fac[1] = 1; for(int i = 2; i <= 8; i++) fac[i] = fac[i - 1] * i; } int getID(int *data) { int id = 0; for(int i = 0; i < 9; i++) { int k = 0; for(int j = i + 1; j < 9; j++) { if(data[j] < data[i]) k++; } id += fac[9 - i - 1] * k; } return id; } void bfs() { node front; memset(isUsed, false, sizeof(isUsed)); memcpy(front.s, tar, sizeof(tar)); front.step = 0; q[0] = front; int id = getID(tar); isUsed[id] = true; ans[id] = 0; int beg = 0, end = 1; while(beg < end) { node tmp; tmp.step = q[beg].step + 1; tmp.s[0] = q[beg].s[1]; tmp.s[1] = q[beg].s[4]; tmp.s[2] = q[beg].s[2]; tmp.s[3] = q[beg].s[0]; tmp.s[4] = q[beg].s[3]; tmp.s[5] = q[beg].s[5]; tmp.s[6] = q[beg].s[6]; tmp.s[7] = q[beg].s[7]; tmp.s[8] = q[beg].s[8]; id = getID(tmp.s); if(!isUsed[id]) { q[end++] = tmp; ans[id] = tmp.step; isUsed[id] = true; } tmp.s[0] = q[beg].s[0]; tmp.s[1] = q[beg].s[2]; tmp.s[2] = q[beg].s[5]; tmp.s[3] = q[beg].s[3]; tmp.s[4] = q[beg].s[1]; tmp.s[5] = q[beg].s[4]; tmp.s[6] = q[beg].s[6]; tmp.s[7] = q[beg].s[7]; tmp.s[8] = q[beg].s[8]; id = getID(tmp.s); if(!isUsed[id]) { q[end++] = tmp; ans[id] = tmp.step; isUsed[id] = true; } tmp.s[0] = q[beg].s[0]; tmp.s[1] = q[beg].s[1]; tmp.s[2] = q[beg].s[2]; tmp.s[3] = q[beg].s[4]; tmp.s[4] = q[beg].s[7]; tmp.s[5] = q[beg].s[5]; tmp.s[6] = q[beg].s[3]; tmp.s[7] = q[beg].s[6]; tmp.s[8] = q[beg].s[8]; id = getID(tmp.s); if(!isUsed[id]) { q[end++] = tmp; ans[id] = tmp.step; isUsed[id] = true; } tmp.s[0] = q[beg].s[0]; tmp.s[1] = q[beg].s[1]; tmp.s[2] = q[beg].s[2]; tmp.s[3] = q[beg].s[3]; tmp.s[4] = q[beg].s[5]; tmp.s[5] = q[beg].s[8]; tmp.s[6] = q[beg].s[6]; tmp.s[7] = q[beg].s[4]; tmp.s[8] = q[beg].s[7]; id = getID(tmp.s); if(!isUsed[id]) { q[end++] = tmp; ans[id] = tmp.step; isUsed[id] = true; } beg++; } } int main() { setFac(); bfs(); while(cin >> init[0]){ for(int i = 1; i < 9; i++) cin >> init[i]; cout << ans[getID(init)] << endl; } return 0; }