UVa 387 - A Puzzling Problem

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題目：俄羅斯方塊、七巧板；已知一些放塊，不能旋轉方塊，問能否拼成4x4的方形。

分析：搜索。dfs枚舉所有放塊，然後模擬所有可能擺放的情況，求解判斷即可。

說明：注意每塊拼圖都要用上╮(╯▽╰)╭。

#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>using namespace std;typedef struct _block{    int  row, column;    int  occupy;    char data[4][5];}block;block B[100];char square[4][5] = {0};int cover(int id, int r, int c){    if (r+B[id].row > 4 || c+B[id].column > 4) return 0;    for (int i = r; i < r+B[id].row; ++ i)        for (int j = c; j < c+B[id].column; ++ j) {            if (square[i][j] != 0 && B[id].data[i-r][j-c] != '0')                return 0;            if (B[id].data[i-r][j-c] != '0')                square[i][j] = '1'+id;        }    return 1;}int uncover(int id){    for (int i = 0; i < 4; ++ i)    for (int j = 0; j < 4; ++ j)        if (square[i][j] == '1'+id)            square[i][j] = 0;}int dfs(int s, int n, int e){    if (e <= 0) {        for (int i = 0; i < 4; ++ i)            puts(square[i]);        return 1;    }    if (s >= n) return 0;    for (int i = s; i < n; ++ i) {        if (B[i].occupy > e) continue;        for (int r = 0; r <= 4; ++ r)            for (int c = 0; c <= 4; ++ c) {                if (cover(i, r, c) && dfs(i+1, n, e-B[i].occupy))                    return 1;                uncover(i);            }    }    return 0;}int main(){    int t, r, c, cases = 0;    while (cin >> t && t) {int sum = 0;        for (int k = 0; k < t; ++ k) {            cin >> B[k].row >> B[k].column;            B[k].occupy = 0;            for (int i = 0; i < B[k].row; ++ i)                 for (int j = 0; j < B[k].column; ++ j) {                    cin >> B[k].data[i][j];                    if (B[k].data[i][j] == '1') {                        B[k].data[i][j] += k;                        B[k].occupy ++;                    }                }            sum += B[k].occupy;        }                if (cases ++) puts("");         memset(square, 0, sizeof(square));        if (sum != 16 || !dfs(0, t, 16))            puts("No solution possible");     }    return 0;}

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