poj 3740 Easy Finding（Dancing Links 精确覆盖）

Easy Finding

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 16128 Accepted: 4321

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 isM, N (M ≤ 16, N ≤ 300). The next M lines every line containsN 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

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http://www.cnblogs.com/grenet/p/3145800.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;}