2016 World Final D Clock Breaking
阿新 • • 發佈:2019-02-19
題意:
給一個LCD燈在一個連續時間內的燈的情況,要你確定哪些燈一定亮,一定暗和不確定。
思路:
先預處理出每個燈泡在24*60分鐘內的狀態,然後用給的時間內不是全亮也不是全暗的燈取匹配,如果沒有一個時間點能夠匹配,就是invalid。然後在考慮全亮或者全暗的燈,如果有一個匹配位置上那個燈是全亮/暗的,那麼那個燈就是不確定的,否則就是壞的。
#include <bits/stdc++.h> using namespace std; #define LL long long #define pii pair<int, int> #define MP make_pair #define inf 0x3f3f3f3f #define mod 1000000007 #define eps 1e-12 #define Pi acos(-1.0) #define N 5050 #define M 200020 #define PB push_back #define MP make_pair #define fi first #define se second int num[11][7] = { 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, }; int dx[] = {0, 1, 4, 6, 4, 1, 3}; int dy[] = {1, 3, 3, 1, 0, 0, 1}; int dx2[] = {0, 2, 5, 6, 5, 2, 3}; int dy2[] = {2, 3, 3, 2, 0, 0, 2}; int f[1440 + 150][4][7]; int g[120][4][7]; char s[120][10][30]; int n; int cnt; int val[N], all[4][7]; char ans[10][30]; void calc(int a[7], int x) { for(int i = 0; i < 7; ++i) a[i] = num[x][i]; } void calc2(int a[7], int i, int x, int y) { for(int k = 0; k < 7; ++k) { a[k] = s[i][x + dx[k]][y + dy[k]] == 'X'; } } void init() { for(int i = 0; i < 24 * 60 + 150; ++i) { int h = i / 60 % 24; int m = i % 60; calc(f[i][0], h / 10 == 0? 10: h / 10); calc(f[i][1], h % 10); calc(f[i][2], m / 10); calc(f[i][3], m % 10); } } bool checkm() { for(int i = 1; i < n; ++i) { if(s[i][2][10] != s[0][2][10]) { return 0; } if(s[i][4][10] != s[0][4][10]) { return 0; } } if(s[0][2][10] == 'X') ans[2][10] = '?'; else ans[2][10] = '0'; if(s[0][4][10] == 'X') ans[4][10] = '?'; else ans[4][10] = '0'; return 1; } int main() { init(); scanf("%d", &n); for(int i = 0; i < n; ++i) { for(int j = 0; j < 7; ++j) { scanf("%s", s[i][j]); } calc2(g[i][0], i, 0, 0); calc2(g[i][1], i, 0, 5); calc2(g[i][2], i, 0, 12); calc2(g[i][3], i, 0, 17); } for(int i = 0; i < 4; ++i) { for(int j = 0; j < 7; ++j) { int t = g[0][i][j], f = 1; for(int k = 1; k < n; ++k) { if(g[k][i][j] != t) { f = 0; break; } } if(!f) ++cnt; else all[i][j] = 1; } } int x = 0, y[4][7], z[4][7]; memset(y, 0, sizeof y); memset(z, 0, sizeof z); for(int i = 0; i < 1440; ++i) { for(int j = 0; j < 4; ++j) { for(int k = 0; k < 7; ++k) { if(all[j][k]) continue; bool ok = 1; for(int u = 0; u < n; ++u) { if(f[i + u][j][k] != g[u][j][k]) { ok = 0; break; } } if(ok) { val[i]++; } } } if(val[i] == cnt) { x++; for(int j = 0; j < 4; ++j) { for(int k = 0; k < 7; ++k) { if(!all[j][k]) continue; bool ok = 1; for(int u = 0; u < n; ++u) { if(f[i + u][j][k] != g[u][j][k]) { ok = 0; break; } } if(ok) { if(g[0][j][k] == 1) y[j][k]++; else z[j][k]++; } } } } } for(int i = 0; i < 7; ++i) { for(int j = 0; j < 21; ++j) { ans[i][j] = '.'; } ans[i][21] = '\0'; } if(x == 0 || !checkm()) { puts("Impossible"); return 0; } for(int i = 0; i < 4; ++i) { for(int j = 0; j < 7; ++j) { int xx = 0, yy = 0; if(i == 1) yy = 5; if(i == 2) yy = 12; if(i == 3) yy = 17; char c; if(!all[i][j]) { c = 'W'; } else if(g[0][i][j] == 1) { if(y[i][j] > 0) c = '?'; else c = '1'; } else { if(z[i][j] > 0) c = '?'; else c = '0'; } ans[xx + dx[j]][yy + dy[j]] = ans[xx + dx2[j]][yy + dy2[j]] = c; } } for(int i = 0; i < 7; ++i) puts(ans[i]); return 0; }