gym/102021/J GCPC18 模拟拼图

拼图"/>

gym/102021/J GCPC18 模拟拼图

模拟拼图

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 << "\n";
#define pb push_back
#define pq priority_queuetypedef 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 '\n'#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\n", 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;
}

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