hihocoder 1054 滑动解锁 dfs

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详细分析见滑动解锁分析

AC代码

#include <cstdio>#include <cmath>#include <cctype>#include <algorithm>#include <cstring>#include <utility>#include <string>#include <iostream>#include <map>#include <set>#include <vector>#include <queue>#include <stack>using namespace std;#pragma comment(linker, "/STACK:1024000000,1024000000") #define eps 1e-10#define inf 0x3f3f3f3f#define PI pair<int, int> typedef long long LL;const int maxn = 10 + 5;int vis[maxn], ans, n, G[maxn][maxn], e[maxn][maxn];struct edge{int x, y;}a[maxn];void init() {memset(G, 0, sizeof(G));for(int i = 1; i <= 7; i+=3) G[i][i+2] = G[i+2][i] = i+1;for(int i = 1; i <= 3; ++i) G[i][i+6] = G[i+6][i] = i+3;G[1][9] = G[9][1] = G[3][7] = G[7][3] = 5;}void dfs(int u, int cnt) {//is it possiblefor(int i = 0; i < n; ++i) {int x = a[i].x, y = a[i].y;if(vis[x] && vis[y] && !e[x][y] && !e[y][x]) return; //一条边的两个端点都被使用，但是这条边没使用则剪枝 }int flag = 1;for(int i = 0; i < n; ++i) {if(!e[a[i].x][a[i].y] && !e[a[i].y][a[i].x]) {flag = 0;break;}}if(flag && cnt >= 4) ans++;for(int i = 1; i <= 9; ++i) {if(u != i && !vis[i] && vis[G[u][i]]) {vis[i] = e[u][i] = 1;dfs(i, cnt+1);vis[i] = e[u][i] = 0;}}}int main() {init();int T;scanf("%d", &T);while(T--) {scanf("%d", &n);for(int i = 0; i < n; ++i) scanf("%d%d", &a[i].x, &a[i].y);ans = 0;for(int i = 1; i <= 9; ++i) {memset(vis, 0, sizeof(vis));memset(e, 0, sizeof(e));vis[0] = vis[i] = 1;dfs(i, 1);}printf("%d\n", ans);}return 0;}

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