POJ3740-Easy Finding

Easy Finding

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 18660 Accepted: 5120

Description

Given a M×N matrix A. A_ij ∈ {0, 1} (0 ≤ i < M, 0 ≤ j < N), could you find some rows that let every cloumn contains and only contains one 1.

Input

There are multiple cases ended by EOF. Test case up to 500.The first line of input is M, N (M ≤ 16, N ≤ 300). The next M lines every line contains N integers separated by space.

Output

For each test case, if you could find it output "Yes, I found it", otherwise output "It is impossible" per line.

Sample Input

3 30 1 00 0 11 0 04 40 0 0 11 0 0 01 1 0 10 1 0 0

Sample Output

Yes, I found itIt is impossible

Source

POJ Monthly Contest - 2009.08.23, MasterLuo

题意：给你一个01矩阵，问能不能选出一些行，使得每一列都有且仅在一行的那一列为1

解题思路：舞蹈链的精确覆盖

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cctype>#include <map>#include <cmath>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;const int maxn = 300005;int n, m, a[20][400];struct DLX{int L[maxn], R[maxn], U[maxn], D[maxn];int row[maxn], col[maxn], ans[maxn], sum[maxn];int n, m, num, cnt;void add(int k, int l, int r, int u, int d, int x, int y){L[k] = l;   R[k] = r;   U[k] = u;D[k] = d;   row[k] = x;  col[k] = y;}void reset(int n, int m){this->n = n;   this->m = m;for (int i = 0; i <= m; i++){add(i, i - 1, i + 1, i, i, 0, i);sum[i] = 0;}L[0] = m, R[m] = 0, cnt = m + 1;}void insert(int x, int y){int temp = cnt - 1;if (row[temp] != x){add(cnt, cnt, cnt, U[y], y, x, y);U[D[cnt]] = cnt; D[U[cnt]] = cnt;}else{add(cnt, temp, R[temp], U[y], y, x, y);R[L[cnt]] = cnt; L[R[cnt]] = cnt;U[D[cnt]] = cnt; D[U[cnt]] = cnt;}sum[y]++, cnt++;}void remove(int k){R[L[k]] = R[k], L[R[k]] = L[k];for (int i = D[k]; i != k; i = D[i])for (int j = R[i]; j != i; j = R[j]){D[U[j]] = D[j];U[D[j]] = U[j];sum[col[j]]--;}}void resume(int k){R[L[k]] = k, L[R[k]] = k;for (int i = D[k]; i != k; i = D[i])for (int j = R[i]; j != i; j = R[j]){D[U[j]] = j;U[D[j]] = j;sum[col[j]]++;}}bool dfs(int k){if (!R[0]) { num = k; return true; }int now = R[0];for (int i = now; i != 0; i = R[i])if (sum[now] > sum[i]) now = i;remove(now);for (int i = D[now]; i != now; i = D[i]){ans[k] = row[i];for (int j = R[i]; j != i; j = R[j]) remove(col[j]);if (dfs(k + 1)) return true;for (int j = L[i]; j != i; j = L[j]) resume(col[j]);}resume(now);return false;}}dlx;int main(){while (~scanf("%d %d", &n, &m)){dlx.reset(n, m);for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++){scanf("%d", &a[i][j]);if (a[i][j]) dlx.insert(i, j);}if (dlx.dfs(0)) printf("Yes, I found it\n");else printf("It is impossible\n");}return 0;}