PKU 3740 Easy Finding(优先队列）【转帖】

 

题意：在M*N的矩阵中，每个元素为０或１,现在问你能否找到若干行，使得每一列只有１个１.

思路：bfs + 优先队列

源代码：

#include <iostream>#include <cstring>#include <cstdio>#include <queue>using namespace std;const int MAXN = 17;const int MAXM = 301;int matrix[MAXN][MAXM];int link[MAXN][MAXN];//判断i行和j行能否组合int rowCnt[MAXN];//rowCnt判断i行中1的总数int m, n;struct node_T { int row[MAXN];//记录已经合并哪些行 int rCnt;//记录行的记录数 int cnt;//记录列中1的总数 bool operator < (const node_T &nod) const//运算符重载 { return cnt < nod.cnt; }};bool bfs() { priority_queue<node_T> Q;//优先级队列 node_T cur, tmp; int i, j; for(i = 0; i < m; i ++) { cur.row[0] = i;//初始化队列，每行入队列 cur.rCnt = 1; cur.cnt = rowCnt[i]; Q.push(cur); } while(! Q.empty()) { cur = Q.top(); Q.pop(); if(cur.cnt == n) return true; tmp = cur; for(i = cur.row[cur.rCnt - 1] + 1; i < m; i ++) { for(j = 0; j < cur.rCnt; j ++) { if(! link[i][cur.row[j]])//两行可以放在一起 break; } if(j == cur.rCnt) { tmp.rCnt = cur.rCnt + 1;//添加一行到结构中，记录在结构中 tmp.row[tmp.rCnt - 1] = i; tmp.cnt = cur.cnt + rowCnt[i]; Q.push(tmp);//加入队列中 } } } return false;}int main() { int i, j, k; while(scanf("%d%d", &m, &n) != EOF) { memset(rowCnt, 0, sizeof(rowCnt)); for(i = 0; i < m; i ++) { for(j = 0; j < n; j ++) { scanf("%d", &matrix[i][j]); rowCnt[i] += matrix[i][j]; } } for(i = 0; i < m; i ++) { link[i][i] = 0; for(j = i + 1; j < m; j ++) { for(k = 0; k < n; k ++) { if(matrix[i][k] & matrix[j][k]) break; } if(k < n) { link[i][j] = link[j][i] = 0; } else { link[i][j] = link[j][i] = 1; } } } if(bfs()) printf("Yes, I found it/n"); else printf("It is impossible/n"); } return(0);}