hdu1195 Open the Lock (BFS)

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题意：给出两个4位的数字，都由1~9组成，要求经过一系列的操作将第一个数字变为第二个数字。操作有两种，

第一种：选择一位加上或减去1，9+1=1，1-1=9。

第二种：交换相邻的两位。第一位和第四位不相邻。

做法：简单bfs

代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <queue>#include <string>using namespace std;struct Node{int stat;int step;Node() {}Node(int a, int b) : stat(a), step(b) {}};int T;int start, end;queue<Node> q;int vis[11111];void push(Node nex) {if (!vis[nex.stat]) {vis[nex.stat] = 1;q.push(nex);}}int bfs() {memset(vis, 0, sizeof(vis));while (!q.empty()) q.pop();Node e = Node(start, 0);vis[start] = 1;q.push(e);while (!q.empty()) {e = q.front();q.pop();if (e.stat == end) return e.step;Node nex;int a = e.stat / 1000;int b = e.stat % 1000 / 100;int c = e.stat % 100 / 10;int d = e.stat % 10;if (a == 9) nex = Node(e.stat - 8000, e.step + 1);else nex = Node(e.stat + 1000, e.step + 1);push(nex);if (a == 1) nex = Node(e.stat + 8000, e.step + 1);else nex = Node(e.stat - 1000, e.step + 1);push(nex);if (b == 9) nex = Node(e.stat - 800, e.step + 1);else nex = Node(e.stat + 100, e.step + 1);push(nex);if (b == 1) nex = Node(e.stat + 800, e.step + 1);else nex = Node(e.stat - 100, e.step + 1);push(nex);if (c == 9) nex = Node(e.stat - 80, e.step + 1);else nex = Node(e.stat + 10, e.step + 1);push(nex);if (c == 1) nex = Node(e.stat + 80, e.step + 1);else nex = Node(e.stat - 10, e.step + 1);push(nex);if (d == 9) nex = Node(e.stat - 8, e.step + 1);else nex = Node(e.stat + 1, e.step + 1);push(nex);if (d == 1) nex = Node(e.stat + 8, e.step + 1);else nex = Node(e.stat - 1, e.step + 1);push(nex);nex = Node(b * 1000 + a * 100 + c * 10 + d, e.step + 1);push(nex);nex = Node(a * 1000 + c * 100 + b * 10 + d, e.step + 1);push(nex);nex = Node(a * 1000 + b * 100 + d * 10 + c, e.step + 1);push(nex);}return 0;}int main() {scanf("%d", &T);while (T--) {scanf("%d%d", &start, &end);int ans = bfs();printf("%d\n", ans);}return 0;}

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