poj 3740 Easy Finding（Dancing Links 精确覆盖）

Easy Finding

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16128		Accepted: 4321

DescriptionGiven a M×N matrix A. Aij ∈ {0, 1} (0 ≤ i < M, 0 ≤ j < N), could you find some rows that let every cloumn contains and only contains one 1.InputThere are multiple cases ended by EOF. Test case up to 500.The first line of input isM, N (M ≤ 16, N ≤ 300). The next M lines every line containsN integers separated by space.OutputFor each test case, if you could find it output "Yes, I found it", otherwise output "It is impossible" per line.Sample Input

3 3
0 1 0
0 0 1
1 0 0
4 4
0 0 0 1
1 0 0 0
1 1 0 1
0 1 0 0

Sample Output

Yes, I found it
It is impossible

题意：给出一个矩阵，问是否存在一些行使得每一列有且只有一个1。

思路：Dancing Links。详见/content/3695284.html

AC代码：

#include <cstdio>#include <cstring>#include <iostream>#include <cmath>#include <algorithm>#include <bitset>#include <queue>#define ll long longusing namespace std;const int maxn = 6005;const int INF = 1e9;int n, m, cnt, head;int L[maxn], R[maxn], U[maxn], D[maxn], S[maxn], C[maxn], H[maxn];inline void add_link(int i, int j){C[++cnt] = j;S[j]++;D[cnt] = j;U[cnt] = U[j];if(H[i]) R[cnt] = H[i], L[cnt] = L[H[i]];else R[cnt] = L[cnt] = cnt;H[i] = cnt;U[D[cnt]] = cnt;D[U[cnt]] = cnt;R[L[cnt]] = cnt;L[R[cnt]] = cnt;}void remove(int c){R[L[c]] = R[c];L[R[c]] = L[c];for(int i = D[c]; i != c; i = D[i])for(int j = R[i]; j != i; j = R[j]){U[D[j]] = U[j];D[U[j]] = D[j];S[C[j]]--;}}void resume(int c){for(int i = U[c]; i != c; i = U[i])for(int j = L[i]; j != i; j = L[j]){D[U[j]] = j;U[D[j]] = j;S[C[j]]++;}L[R[c]] = R[L[c]] = c;}bool dance(){if(R[head] == head){puts("Yes, I found it");return true;}int s = INF, c;for(int i = R[head]; i != head; i = R[i])if(S[i] < s) s = S[c = i];remove(c);for(int i = D[c]; i != c; i = D[i]){for(int j = R[i]; j != i; j = R[j]) remove(C[j]);if(dance()) return true;for(int j = L[i]; j != i; j = L[j]) resume(C[j]);}resume(c);return false;}int main(){int c;head = 0;while(~scanf("%d%d", &n, &m)){cnt = m;for(int i =- 0; i <= m; i++){C[i] = U[i] = D[i] = i;L[i + 1] = i;R[i] = i + 1;S[i] = 0;}L[0] = m, R[m] = 0;for(int i = 1; i <= n; i++){H[i] = 0;for(int j = 1; j <= m; j++){c = getchar();while(!isdigit(c)) c = getchar();if(c == '1') add_link(i, j);}}if (!dance()) puts("It is impossible");}return 0;}

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