uva 687 - Lattice Practices（暴力)

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题目链接：uva 687 - Lattice Practices

题目大意：给出十个拼图，输出有多少种组合方案，旋转翻转后相同的算一种。

解题思路：被题目坑了，虽然说给的拼图不会有相同的，但是说平涂可以调转使用，调转后可能就相同了。所有用set将已经找到的组成放法的所有旋转方式全部记录。

#include <stdio.h>#include <string.h>#include <string>#include <set>#include <algorithm>#include <iostream>using namespace std;const int N = 20;const int M = 50;int ans, g[N];set<string> rec;int d[M], c[M];char sta[M][N];void handle() {for (int i = 0; i < 32; i++) {int k = i, t = 0;for (int j = 0; j < 5; j++) {int m = k % 2; k /= 2;t = t * 2 + m;sta[i][4-j] = '0' + m;}d[i] = t;}}void rechange(int id, char* str) {int t = 0, k = 0;for (int i = 0; i < 5; i++) {t = t * 2 + str[i] - '0';k = k * 2 + str[4-i] - '0';}c[t]++; c[k]++;}bool init() {rec.clear();ans = 0;memset(c, 0, sizeof(c));char str[N];for (int i = 0; i < 10; i++) {scanf("%s", str);if (strcmp(str, "END") == 0) return false;rechange(i, str); }return true;}void add(char (*x)[N], char (*y)[N]) {string str;for (int i = 0; i < 5; i++) str = str + x[i];for (int i = 0; i < 5; i++) str = str + y[i];rec.insert(str);}void ret(char(*x)[N]) {char y[N][N];for (int i = 0; i < 5; i++) strcpy(y[i], x[i]);for (int i = 0; i < 5; i++) {for (int j = 0; j < 5; j++) x[j][4-i] = y[i][j];}}void del() {ans++;char x[N][N], y[N][N];for (int i = 0; i < 5; i++) {strcpy(x[i], sta[g[i]]);strcpy(y[i], sta[g[i+5]]);}for (int i = 0; i < 4; i++) {add(x, y);ret(x); ret(y);}for (int i = 0; i < 5; i++) {strcpy(x[i], sta[g[4-i]]);strcpy(y[i], sta[d[g[i+5]]]);}for (int i = 0; i < 4; i++) {add(x, y);ret(x); ret(y);}}bool judge() {int r[M];memset(r, 0, sizeof(r));for (int i = 0; i < 5; i++) {int k = 0;for (int j = 0; j < 5; j++) {k = k * 2 + ( g[j] & (1<<i) ? 0 : 1 );}if (r[k] >= c[k]) return false;g[i+5] = k;r[k]++; r[d[k]]++;}string str;for (int i = 0; i < 10; i++) {str = str + sta[g[i]];}if (rec.find(str) != rec.end()) return false;return true;}void dfs(int deep) {if (deep >= 5) {if (judge()) del();return;}for (int i = 0; i < 32; i++) {if (c[i] == 0) continue;c[i]--; c[d[i]]--;g[deep] = i; dfs(deep + 1);c[i]++; c[d[i]]++;}}int main() {handle();while (init()) {dfs(0);printf("%d\n", ans / 2);}return 0;}

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