Friday, November 27, 2015

[itint5]直角路线遍历棋盘 - 阿牧遥 - 博客园


[itint5]直角路线遍历棋盘 - 阿牧遥 - 博客园
https://github.com/zdzapple/itint5/blob/master/%E7%9B%B4%E8%A7%92%E8%B7%AF%E7%BA%BF%E9%81%8D%E5%8E%86%E6%A3%8B%E7%9B%98.cpp
无向图的欧拉路径:只有0或者2个度数为奇数的点
欧拉回路:每个顶点的度数都是偶数
有向图的欧拉路径:起始顶点的出度(从这个顶点发出的有向边的数量)比入度(指向这个顶点的有向边的数量)多1,结束顶点的出度比入度少1,而其它顶点的出度和入度都相等。
欧拉回路:以下两个之一:
每个顶点的出度和入度都相等;
存在一系列的(有向)环,使得图里的每一条边都恰好属于某一个环。

具体的转换方式为:n,m的棋盘,建一个包含n+m个顶点的图G(为了方便说明,类似二分图将其分为两列,左边n个顶点,右边m个顶点,分别代表n行和n列)。对于目标格子(i,j),左边第i个顶点和右边第j个顶点连一条边。最后的问题其实就是问转换之后的图G是否存在欧拉欧拉回路或者欧拉路径。
证明:相邻两步为直角,其实就是从某一行变到某一列。访问图G中的一条边,意味着访问棋盘中的一个目标点。由于图G中的边只连接左边的点(代表某一行)和右边的点(代表某一列),因此访问一条边就意味着从某一行变到了某一列,也就是转直角了。
所以问题变为能否从一点出发访问G中的所有边有且仅有一次。这个就是欧拉回路问题了。

//如果存在满足条件的遍历,返回true,否则返回false
bool existPath(vector<int> &x, vector<int> &y)
{
int n = x.size();
if (x.empty())
return true;
int rowNum = x[0];
int colNum = y[0];
int i, j;
for (i = 0; i < n; ++  i)
{
if (rowNum < x[i])
rowNum = x[i];
if (colNum < y[i])
colNum = y[i];
}
// the graoh should be connected graph
//判断连通性
    bool xa[50] = {false};
    bool ya[50] = {false};
    xa[x[0]] = true;
    ya[y[0]] = true;
    i = 1, j = x.size() - 1;
    while (i <= j)
{
        if(xa[x[i]] || ya[y[i]]) {
            xa[x[i]] = true;
            ya[y[i]] = true;
            i ++;
            j = x.size() - 1;
        }
        else {
            int xt = x[j], yt = y[j];
            x[j] = x[i];
            y[j] = y[i];
            x[i] = xt;
            y[i]= yt;
            j--;
        }
    }
    if (i!=x.size())
return false;


// the number of odd degresses should be 0 or 2
vector<int> leftDegress(rowNum + 1, 0);
vector<int> rightDegress(colNum + 1, 0);
for (i = 0; i < n; ++ i)
{
leftDegress[x[i]] ++;
rightDegress[y[i]] ++;
}
int oddDegressCount = 0;
for (i = 0; i <= rowNum; ++ i)
{
if (leftDegress[i] % 2)
oddDegressCount ++;
}
for (i = 0; i <= colNum; ++ i)
{
if (rightDegress[i] % 2)
oddDegressCount ++;
}
return oddDegressCount == 0 || oddDegressCount == 2;
}

这题一开始直接用暴力的DFS来做,果然到25的规模就挂了.
vector<bool> visited(50, false);
vector<vector<int> > vec_row(50);
vector<vector<int> > vec_col(50);
 
bool findPath(vector<int> &x, vector<int> &y, int idx, int depth, int direction) {
    if (depth == x.size()) return true;
    visited[idx] = true;
    if (direction == 0) {
        int row = x[idx];
        for (int i = 0; i < vec_row[row].size(); i++) {
            if (!visited[vec_row[row][i]]) {
                if (findPath(x, y, vec_row[row][i], depth+1, 1))
                    return true;
            }
        }
    else {
        int col = y[idx];
        for (int i = 0; i < vec_col[col].size(); i++) {
            if (!visited[vec_col[col][i]]) {
                if (findPath(x, y, vec_col[col][i], depth+1, 0))
                    return true;
            }
        }
    }
     
    visited[idx] = false;
    return false;
}
 
 
//如果存在满足条件的遍历,返回true,否则返回false
bool existPath(vector<int> &x, vector<int> &y) {
    int k = x.size();
    if (k == 0) return true;
    for (int i = 0; i < k; i++) {
        vec_row[x[i]].push_back(i);
        vec_col[y[i]].push_back(i);
    }
    for (int i = 0; i < k; i++) {
        if (findPath(x, y, i, 1, 0) ||
            findPath(x, y, i, 1, 1)) {
                return true;
            }
    }
    return false;
}

这题可以直接转换为欧拉回路(路径)问题,这样,如果有解的时候要输出遍历路径的时候,也比较好办了。
具体的转换方式为:n,m的棋盘,建一个包含n+m个顶点的图G(为了方便说明,类似二分图将其分为两列,左边n个顶点,右边m个顶点,分别代表n行和n列)。对于目标格子(i,j),左边第i个顶点和右边第j个顶点连一条边。最后的问题其实就是问转换之后的图G是否存在欧拉欧拉回路或者欧拉路径。
证明:相邻两步为直角,其实就是从某一行变到某一列。访问图G中的一条边,意味着访问棋盘中的一个目标点。由于图G中的边只连接左边的点(代表某一行)和右边的点(代表某一列),因此访问一条边就意味着从某一行变到了某一列,也就是转直角了。
所以问题变为能否从一点出发访问G中的所有边有且仅有一次。这个就是欧拉回路问题了。
 所以欧拉路径是:1.连通;2.奇点为2,为0时是欧拉回路。
 这里的连通我用并查集来做。注意写并查集的merge时,要先找到根,再merge。
vector<int> djset;
 
int find(int i) {
    int x = i;
    while (djset[x] != x) {
        x = djset[x];
    }
    djset[i] = x;
    return x;
}
 
void merge(int i, int j) {
    djset[find(i)] = djset[find(j)];
}
 
//如果存在满足条件的遍历,返回true,否则返回false
bool existPath(vector<int> &x, vector<int> &y) {
    vector<int> axis(100, 0);
    // 计算奇点数目
    for (int i = 0; i < x.size(); i++) {
        axis[x[i]] = !axis[x[i]]; // 奇偶变换
    }
    for (int i = 0; i < y.size(); i++) {
        axis[y[i]+50] = !axis[y[i]+50];
    }
    int count = 0;
    for (int i = 0; i < axis.size(); i++) {
        if (axis[i]) count++;
    }
    if (count != 0 && count != 2) return false;
     
    djset.resize(x.size());
    for (int i = 0; i < djset.size(); i++) {
        djset[i] = i;
    }
     
    // 判断连通性
    for (int i = 0; i < x.size(); i++) {
        for (int j = i+1; j < x.size(); j++) {
            if (x[i] == x[j]) {
                merge(i, j);
            }
        }
    }
    for (int i = 0; i < y.size(); i++) {
        for (int j = i+1; j < y.size(); j++) {
            if (y[i] == y[j]) {
                merge(i, j);
            }
        }
    }
     
    for (int i = 0; i < x.size(); i++) {
        if (find(i) != find(0)) {
            return false;
        }
    }
    return true;
}
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