POJ 3740 - Easy Finding （Dancing links）

最新推荐文章于 2014-10-16 16:57:07 发布

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最新推荐文章于 2014-10-16 16:57:07 发布

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本文探讨了深度学习技术在音视频处理、AR特效、AI音视频处理等领域的应用，包括图像处理、音视频编解码、自然语言处理、区块链与隐私计算等关键技术。介绍了从图像色彩空间到AR特效，从语音识别到AI边缘计算的最新进展。

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Easy Finding

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15738		Accepted: 4177

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

Sample Output

Yes, I found it
It is impossible

Source

POJ Monthly Contest - 2009.08.23, MasterLuo

题意：

给一个n*m的01矩阵，问能否找到k行，使得每一列刚好只有一行为1

Dancing links 点击打开链接

#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <queue>
#include <set>

using namespace std;

//#define WIN
#ifdef WIN
typedef __int64 LL;
#define iform "%I64d"
#define oform "%I64d\n"
#else
typedef long long LL;
#define iform "%lld"
#define oform "%lld\n"
#endif

#define S64I(a) scanf(iform, &(a))
#define P64I(a) printf(oform, (a))
#define REP(i, n) for(int (i)=0; (i)<n; (i)++)
#define REP1(i, n) for(int (i)=1; (i)<=(n); (i)++)
#define FOR(i, s, t) for(int (i)=(s); (i)<=(t); (i)++)

const int INF = 0x3f3f3f3f;
const double eps = 10e-9;
const double PI = (4.0*atan(1.0));

const int maxn = 6000 + 20;
int U[maxn], D[maxn], L[maxn], R[maxn], S[maxn];
int H[maxn];
int COL[maxn]; // 列号
int n, m;
int size;

void init() {
    for(int i=0; i<=m; i++) {
        U[i] = D[i] = i;
        L[i] = i-1;
        R[i] = i+1;
        S[i] = 0;
    }
    R[m] = 0;
    L[0] = m;
    size = m+1;
    memset(H, -1, sizeof(H));
}

void insert(int row, int col) {
    U[size] = U[col];
    D[size] = col;
    D[U[col]] = size;
    U[col] = size;
    if(H[row] == -1) {
        H[row] = R[size] = L[size] = size;
    } else {
        int rh = H[row];
        R[size] = rh;
        L[size] = L[rh];
        R[L[rh]] = size;
        L[rh] = size;
    }
    S[col]++;
    COL[size] = col;
    size++;
}

void remove(int c) {
    L[R[c]] = L[c];
    R[L[c]] = R[c];
    for(int i=D[c]; i!=c; i=D[i]) {
        for(int j=R[i]; j!=i; j=R[j]) {
            S[COL[j]]--;
            U[D[j]] = U[j];
            D[U[j]] = D[j];
        }
    }
}

void resume(int c) {
    L[R[c]] = c;
    R[L[c]] = c;
    for(int i=D[c]; i!=c; i=D[i]) {
        for(int j=L[i]; j!=i; j=L[j]) {
            S[COL[j]]++;
            U[D[j]] = j;
            D[U[j]] = j;
        }
    }
}

bool dfs() {
    if(R[0] == 0) {
        return true;
    }
    int c = R[0];
    int mins = S[c];
    for(int i=R[0]; i!=0; i=R[i]) if(S[i] < mins) {
        c = i;
        mins = S[i];
    }
    remove(c);
    for(int i=D[c]; i!=c; i=D[i]) {
        for(int j=R[i]; j!=i; j=R[j]) remove(COL[j]);
        if(dfs()) return true;
        for(int j=L[i]; j!=i; j=L[j]) resume(COL[j]);
    }
    resume(c);
    return false;
}

int main() {

    while(scanf("%d%d", &n, &m) != EOF) {
        init();
        REP1(i, n) REP1(j, m) {
            int t;
            scanf("%d", &t);
            if(t) insert(i, j);
        }
        if(dfs()) {
            puts("Yes, I found it");
        } else {
            puts("It is impossible");
        }
    }

    return 0;
}