题意：
给定n块拼图，每个拼图为四方形，对应四条边有四个数字，如果为0，表示这个边是在边界的，其他数字表示和另一个拼图的一条边相接。保证每个非零数只出现两次。
思路：
模拟，但是要注意几个情况，第一就是只有一行或一列的时候，对于0的判断，还有就是拼图的个数要和n相等。
#include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> using namespace std; #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << " "; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; //typedef __int128 bll; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int,pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl ' ' #define OKC ios::sync_with_stdio(false);cin.tie(0) #define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行 #define REP(i , j , k) for(int i = j ; i < k ; ++i) #define max3(a,b,c) max(max(a,b), c); #define min3(a,b,c) min(min(a,b), c); //priority_queue<int ,vector<int>, greater<int> >que; const ll mos = 0x7FFFFFFF; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 const int mod = 9999973; const double esp = 1e-8; const double PI=acos(-1.0); const double PHI=0.61803399; //黄金分割点 const double tPHI=0.38196601; template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x=f?-x:x; } /*-----------------------showtime----------------------*/ const int maxn = 3e5+9; int a[maxn][10]; int mp[maxn*4][2],vis[maxn]; vector<int>res[maxn]; int get(int id,int nd1,int nd2){ int res = -1; for(int i=0; i<=3; i++){ if(a[id][i] == nd1 && a[id][i+1] == nd2) { res = i; } } return res; } int main(){ int n; scanf("%d", &n); int flag =1, st = -1; for(int i=1; i<=n; i++){ scanf("%d%d%d%d", &a[i][0], &a[i][1], &a[i][2], &a[i][3]); a[i][4] = a[i][0]; a[i][5] = a[i][1]; a[i][6] = a[i][2]; a[i][7] = a[i][3]; int cnt = 0; for(int j=0; j<=3; j++){ if(a[i][j] == 0) cnt++; else { if(mp[a[i][j]][0] == 0) mp[a[i][j]][0] = i; else if(mp[a[i][j]][1] == 0)mp[a[i][j]][1] = i; // else { puts("impossible");return 0;} } } if(cnt >= 2 && st == -1){ int tmp = get(i, 0, 0); // debug(tmp); if(tmp >=0) st = i,a[st][8] = tmp; } } if(st == -1) { puts("impossible"); return 0; } int h=0,w=0; w = 0; res[0].pb(st); vis[st] = 1; for(;;){ int id = res[0][w]; int num = a[id][a[id][8] + 3]; if(num == 0) break; int nx = -1; if(vis[mp[num][0]] == 0) nx = mp[num][0]; else if(vis[mp[num][1]] == 0) nx = mp[num][1]; if(nx == -1) {puts("impossible");return 0;} int tmp = get(nx, 0, num); if(tmp == -1) {puts("impossible");return 0;} a[nx][8] = tmp; res[0].pb(nx); w++; vis[nx] = 1; } for(; ;){ int id = res[h][0]; int num = a[id][a[id][8] + 2]; // debug(num); if(num == 0) break; int nx = -1; if(vis[mp[num][0]] == 0) nx = mp[num][0]; else if(vis[mp[num][1]] == 0) nx = mp[num][1]; if(nx == -1) {puts("impossible");return 0;} int tmp = get(nx, num, 0); // debug(nx); if(tmp == -1) {puts("impossible");return 0;} a[nx][8] = tmp; h++; res[h].pb(nx); vis[nx] = 1; } // cout<<h<<" "<<w<<endl; for(int i=1; i<=h; i++){ for(int j=1; j<=w; j++){ int num1id = res[i-1][j]; int num2id = res[i][j-1]; int num1 = a[num1id][a[num1id][8] + 2]; int num2 = a[num2id][a[num2id][8] + 3]; // cout<<num1id<<" "<<num2id<<endl; // cout<<num1<<" "<<num2<<endl; if(num1 == 0 || num2 == 0) {puts("impossible"); return 0;} int nx1 = -1, nx2 = -1; if(vis[mp[num1][0]] == 0) nx1 = mp[num1][0]; else if(vis[mp[num1][1]] == 0) nx1 = mp[num1][1]; if(vis[mp[num2][0]] == 0) nx2= mp[num2][0]; else if(vis[mp[num2][1]] == 0) nx2 = mp[num2][1]; if(nx1 == -1 || nx2 == -1) {puts("impossible");return 0;} if(nx1 != nx2) {puts("impossible");return 0;} int tmp = get(nx1, num1, num2); if(tmp == -1) {puts("impossible");return 0;} a[nx1][8] = tmp; res[i].pb(nx1); vis[nx1] = 1; if(i==h) if(a[nx1][tmp + 2] != 0) {puts("impossible");return 0;} if(j==w) if(a[nx1][tmp + 3] != 0) {puts("impossible");return 0;} } } if((h+1) * (w + 1) != n) {puts("impossible");return 0;} if(h == 0) { for(int i=0; i<=w; i++) { int id = res[0][i]; int t = a[id][8]; if(a[id][t+2] != 0) {puts("impossible");return 0;} if(i == w && a[id][t+3] != 0) {puts("impossible");return 0;} } } if(w == 0){ for(int i=0; i<=h; i++){ int id = res[i][0]; int t = a[id][8]; if(a[id][t+3] != 0) {puts("impossible");return 0;} if(i == h && a[id][t+2] != 0) {puts("impossible");return 0;} } } printf("%d %d ", h+1, w+1); for(int i=0; i<=h; i++){ for(int j=0; j<=w; j++){ printf("%d ", res[i][j]); } puts(""); } return 0; }