【01BFS】# P4667 [BalticOI 2011] Switch the Lamp On 电路维修 (Day1)|普及+

本文涉及知识点

C++BFS算法

题目描述

Casper is designing an electronic circuit on a $\times M$ rectangular grid plate. There are $\times M$ square tiles that are aligned to the grid on the plate. Two (out of four) opposite corners of each tile are connected by a wire.

A power source is connected to the top left corner of the plate. A lamp is connected to the bottom right corner of the plate. The lamp is on only if there is a path of wires connecting power source to lamp. In order to switch the lamp on, any number of tiles can be turned by 90° (in both directions).
在这里插入图片描述

在这里插入图片描述

In the picture above the lamp is off. If any one of the tiles in the second column from the right is turned by 90° , power source and lamp get connected, and the lamp is on.

Write a program to find out the minimal number of tiles that have to be turned by 90° to switch the lamp on.

输入格式

The first line of input contains two integer numbers $N$ and $M$ , the dimensions of the plate. In each of the following $N$ lines there are $M$ symbols – either \ or / – which indicate the direction of the wire connecting the opposite vertices of the corresponding tile.

输出格式

There must be exactly one line of output. If it is possible to switch the lamp on, this line must contain only one integer number: the minimal number of tiles that have to be turned to switch on the lamp. If it is not possible, output the string: NO SOLUTION

输入输出样例 #1

输入 #1

3 5
\\/\\
\\///
/\\\\

输出 #1

说明/提示

对于 $40\%$ 的数据， $\le N \le 4$ ， $\le M \le 5$ 。

对于所有数据， $\le N,M \le 500$ 。

BFS 双向队列

一，本题是点图，不是单格图。故：r $\in$ [0,N],c $\in$ [0,M]。
二，本题能否保证每个点都只访问一次？vis记录最少改变次数，改变次数变少，才需要重新处理当前点的后续。因为是01BFS，所以可以保证所有节点只访问一次。
三，用双向队列，如果不改变入队首，改变入队尾。这样优先处理不改变，可以提速。
时间复杂度：O（NM）
任意时刻：双向队列要么全部是x，要么前边是x，后边是x+1。
x入队首，x+1入队尾，扔保持这种状态。故x处理完之前，不会处理x+1。

代码

核心代码

#include <iostream>
#include <sstream>
#include <vector>
#include<map>
#include<unordered_map>
#include<set>
#include<unordered_set>
#include<string>
#include<algorithm>
#include<functional>
#include<queue>
#include <stack>
#include<iomanip>
#include<numeric>
#include <math.h>
#include <climits>
#include<assert.h>
#include<cstring>
#include<list>#include <bitset>
using namespace std;template<class T1, class T2>
std::istream& operator >> (std::istream& in, pair<T1, T2>& pr) {in >> pr.first >> pr.second;return in;
}template<class T1, class T2, class T3 >
std::istream& operator >> (std::istream& in, tuple<T1, T2, T3>& t) {in >> get<0>(t) >> get<1>(t) >> get<2>(t) ;return in;
}template<class T1, class T2, class T3, class T4 >
std::istream& operator >> (std::istream& in, tuple<T1, T2, T3, T4>& t) {in >> get<0>(t) >> get<1>(t) >> get<2>(t) >> get<3>(t);return in;
}template<class T = int>
vector<T> Read() {int n;scanf("%d", &n);vector<T> ret(n);for(int i=0;i < n ;i++) {cin >> ret[i];}return ret;
}template<class T = int>
vector<T> Read(int n) {vector<T> ret(n);for (int i = 0; i < n; i++) {cin >> ret[i];}return ret;
}template<int N = 12 * 1'000'000>
class COutBuff
{
public:COutBuff() {m_p = puffer;}template<class T>void write(T x) {int num[28], sp = 0;if (x < 0)*m_p++ = '-', x = -x;if (!x)*m_p++ = 48;while (x)num[++sp] = x % 10, x /= 10;while (sp)*m_p++ = num[sp--] + 48;}inline void write(char ch){*m_p++ = ch;}inline void ToFile() {fwrite(puffer, 1, m_p - puffer, stdout);}
private:char  puffer[N], * m_p;
};template<int N = 12 * 1'000'000>
class CInBuff
{
public:inline CInBuff() {fread(buffer, 1, N, stdin);}inline int Read() {int x(0), f(0);while (!isdigit(*S))f |= (*S++ == '-');while (isdigit(*S))x = (x << 1) + (x << 3) + (*S++ ^ 48);return f ? -x : x;}
private:char buffer[N], * S = buffer;
};template<class TSave, class TRecord >
class CRangUpdateLineTree
{
protected:virtual void OnQuery(const TSave& save, const int& iSaveLeft, const int& iSaveRight) = 0;virtual void OnUpdate(TSave& save, const int& iSaveLeft, const int& iSaveRight, const TRecord& update) = 0;virtual void OnUpdateParent(TSave& par, const TSave& left, const TSave& r, const int& iSaveLeft, const int& iSaveRight) = 0;virtual void OnUpdateRecord(TRecord& old, const TRecord& newRecord) = 0;
};template<class TSave, class TRecord >
class CVectorRangeUpdateLineTree : public CRangUpdateLineTree<TSave, TRecord>
{
public:CVectorRangeUpdateLineTree(int iEleSize, TSave tDefault, TRecord tRecordNull) :m_iEleSize(iEleSize), m_save(iEleSize * 4, tDefault), m_record(iEleSize * 4, tRecordNull) {m_recordNull = tRecordNull;}void Update(int iLeftIndex, int iRightIndex, TRecord value){Update(1, 0, m_iEleSize - 1, iLeftIndex, iRightIndex, value);}void Query(int leftIndex, int rightIndex) {Query(1, 0, m_iEleSize - 1, leftIndex, rightIndex);}//void Init() {//	Init(1, 0, m_iEleSize - 1);//}TSave QueryAll() {return m_save[1];}void swap(CVectorRangeUpdateLineTree<TSave, TRecord>& other) {m_save.swap(other.m_save);m_record.swap(other.m_record);std::swap(m_recordNull, other.m_recordNull);assert(m_iEleSize == other.m_iEleSize);}
protected://void Init(int iNodeNO, int iSaveLeft, int iSaveRight)//{//	if (iSaveLeft == iSaveRight) {//		this->OnInit(m_save[iNodeNO], iSaveLeft);//		return;//	}//	const int mid = iSaveLeft + (iSaveRight - iSaveLeft) / 2;//	Init(iNodeNO * 2, iSaveLeft, mid);//	Init(iNodeNO * 2 + 1, mid + 1, iSaveRight);//	this->OnUpdateParent(m_save[iNodeNO], m_save[iNodeNO * 2], m_save[iNodeNO * 2 + 1], iSaveLeft, iSaveRight);//}void Query(int iNodeNO, int iSaveLeft, int iSaveRight, int iQueryLeft, int iQueryRight) {if ((iSaveLeft >= iQueryLeft) && (iSaveRight <= iQueryRight)) {this->OnQuery(m_save[iNodeNO], iSaveLeft, iSaveRight);return;}if (iSaveLeft == iSaveRight) {//没有子节点return;}Fresh(iNodeNO, iSaveLeft, iSaveRight);const int mid = iSaveLeft + (iSaveRight - iSaveLeft) / 2;if (mid >= iQueryLeft) {Query(iNodeNO * 2, iSaveLeft, mid, iQueryLeft, iQueryRight);}if (mid + 1 <= iQueryRight) {Query(iNodeNO * 2 + 1, mid + 1, iSaveRight, iQueryLeft, iQueryRight);}}void Update(int iNode, int iSaveLeft, int iSaveRight, int iOpeLeft, int iOpeRight, TRecord value){if ((iOpeLeft <= iSaveLeft) && (iOpeRight >= iSaveRight)){this->OnUpdate(m_save[iNode], iSaveLeft, iSaveRight, value);this->OnUpdateRecord(m_record[iNode], value);return;}if (iSaveLeft == iSaveRight) {return;//没有子节点}Fresh(iNode, iSaveLeft, iSaveRight);const int iMid = iSaveLeft + (iSaveRight - iSaveLeft) / 2;if (iMid >= iOpeLeft){Update(iNode * 2, iSaveLeft, iMid, iOpeLeft, iOpeRight, value);}if (iMid + 1 <= iOpeRight){Update(iNode * 2 + 1, iMid + 1, iSaveRight, iOpeLeft, iOpeRight, value);}// 如果有后代，至少两个后代this->OnUpdateParent(m_save[iNode], m_save[iNode * 2], m_save[iNode * 2 + 1], iSaveLeft, iSaveRight);}void Fresh(int iNode, int iDataLeft, int iDataRight){if (m_recordNull == m_record[iNode]){return;}const int iMid = iDataLeft + (iDataRight - iDataLeft) / 2;Update(iNode * 2, iDataLeft, iMid, iDataLeft, iMid, m_record[iNode]);Update(iNode * 2 + 1, iMid + 1, iDataRight, iMid + 1, iDataRight, m_record[iNode]);m_record[iNode] = m_recordNull;}vector<TSave> m_save;vector<TRecord> m_record;TRecord m_recordNull;const int m_iEleSize;
};class C01BFSDis
{
public:C01BFSDis(vector<vector<int>>& vNeiB0, vector<vector<int>>& vNeiB1, int s){m_vDis.assign(vNeiB0.size(), -1);std::deque<std::pair<int, int>> que;que.emplace_back(s, 0);while (que.size()){auto it = que.front();const int cur = it.first;const int dis = it.second;que.pop_front();if (-1 != m_vDis[cur]){continue;}m_vDis[cur] = it.second;for (const auto next : vNeiB0[cur]){if (-1 != m_vDis[next]){continue;}que.emplace_front(next, dis);}for (const auto next : vNeiB1[cur]){if (-1 != m_vDis[next]){continue;}que.emplace_back(next, dis + 1);}}}
public:vector<int> m_vDis;
};class Solution {
public:int Ans(vector<string>& mat) {const int R = mat.size();const int C = mat[0].size();auto Mask = [&](int r, int c) {return (C + 1) * r + c;};const int iMC = (R + 1) * (C + 1);vector<vector<int>> neiBo0(iMC), neiBo1(iMC);for (int r = 0; r < R; r++) {for (int c = 0; c < C; c++) {const int mask1 = Mask(r, c);const int mask2 = Mask(r + 1, c + 1);const int mask3 = Mask(r + 1, c);const int mask4 = Mask(r, c + 1);if ('\\' == mat[r][c]) {neiBo0[mask1].emplace_back(mask2);neiBo0[mask2].emplace_back(mask1);neiBo1[mask3].emplace_back(mask4);neiBo1[mask4].emplace_back(mask3);}else {neiBo0[mask3].emplace_back(mask4);neiBo0[mask4].emplace_back(mask3);neiBo1[mask1].emplace_back(mask2);neiBo1[mask2].emplace_back(mask1);}}}C01BFSDis bfs(neiBo0, neiBo1, 0);return bfs.m_vDis.back();}};int main() {
#ifdef _DEBUGfreopen("a.in", "r", stdin);
#endif // DEBUG	int R, C;cin >> R >> C;vector<string> mat(R);for (int r = 0; r < R; r++) {cin >> mat[r];}auto res = Solution().Ans(mat);if (res < 0) {cout << "NO SOLUTION";}else {cout << res;}	
#ifdef _DEBUG		/*printf("T=%d,", T);*/Out(mat, "mat=");
#endif // DEBUG	return 0;
}

单元测试

vector<string> mat;TEST_METHOD(TestMethod11){mat = { {"\\/"} };auto res = Solution().Ans(mat);AssertEx(-1, res);}TEST_METHOD(TestMethod12){mat = { {"\\/\\"} };auto res = Solution().Ans(mat);AssertEx(0, res);}TEST_METHOD(TestMethod13){mat = { {"///"} };auto res = Solution().Ans(mat);AssertEx(2, res);}TEST_METHOD(TestMethod14){mat = { "\\\\/\\\\","\\\\///","/\\\\\\\\" };auto res = Solution().Ans(mat);AssertEx(1, res);}

扩展阅读

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子墨子言之：事无终始，无务多业。也就是我们常说的专业的人做专业的事。
如果程序是一条龙，那算法就是他的是睛
失败+反思=成功成功+反思=成功