LeetCode1575统计所有可行路径

给你一个 互不相同 的整数数组，其中 locations[i] 表示第 i 个城市的位置。同时给你 start，finish 和 fuel 分别表示出发城市、目的地城市和你初始拥有的汽油总量每一步中，如果你在城市 i ，你可以选择任意一个城市 j ，满足 j != i 且 0 <= j < locations.length ，并移动到城市 j 。从城市 i 移动到 j 消耗的汽油量为 |locations[i] - locations[j]|，|x| 表示 x 的绝对值。
请注意， fuel 任何时刻都 不能 为负，且你 可以 经过任意城市超过一次（包括 start 和 finish ）。
请你返回从 start 到 finish 所有可能路径的数目。
由于答案可能很大， 请将它对 10^9 + 7 取余后返回。
示例 1：
输入：locations = [2,3,6,8,4], start = 1, finish = 3, fuel = 5
输出：4
解释：以下为所有可能路径，每一条都用了 5 单位的汽油：
1 -> 3
1 -> 2 -> 3
1 -> 4 -> 3
1 -> 4 -> 2 -> 3
示例 2：
输入：locations = [4,3,1], start = 1, finish = 0, fuel = 6
输出：5
解释：以下为所有可能的路径：
1 -> 0，使用汽油量为 fuel = 1
1 -> 2 -> 0，使用汽油量为 fuel = 5
1 -> 2 -> 1 -> 0，使用汽油量为 fuel = 5
1 -> 0 -> 1 -> 0，使用汽油量为 fuel = 3
1 -> 0 -> 1 -> 0 -> 1 -> 0，使用汽油量为 fuel = 5
示例 3：
输入：locations = [5,2,1], start = 0, finish = 2, fuel = 3
输出：0
解释：没有办法只用 3 单位的汽油从 0 到达 2 。因为最短路径需要 4 单位的汽油。
提示：
2 <= locations.length <= 100
1 <= locations[i] <= 10⁹
所有 locations 中的整数 互不相同 。
0 <= start, finish < locations.length
1 <= fuel <= 200

动态规划

令n = locations.length

动态规划的状态表示

dp[l][f] 表示 使用f单位的汽油，终点是l的可行路径数。状态数共：nm，故空间复杂度:O(nm)。

动态规划的转移方程

通过前置条件转移后置条件。
枚举符合以下条件的l1：
l1!=l
f1 = |loc[l1]-[l]| + f <= fuel。
dp[l1][[f1] += dp[l][f]
时间复杂度: O(nnm)

动态规划的填表顺序

f从小到大，确保动态规划的无后效性。

动态规划的初始状态

dp[start][0] =1 ，其它全部为0。

动态规划的返回值

dp[finish]之和

代码

核心代码

template<int MOD = 1000000007>
class C1097Int
{
public:
	C1097Int(long long llData = 0) :m_iData(llData% MOD)
	{

	}
	C1097Int  operator+(const C1097Int& o)const
	{
		return C1097Int(((long long)m_iData + o.m_iData) % MOD);
	}
	C1097Int& operator+=(const C1097Int& o)
	{
		m_iData = ((long long)m_iData + o.m_iData) % MOD;
		return *this;
	}
	C1097Int& operator-=(const C1097Int& o)
	{
		m_iData = (m_iData + MOD - o.m_iData) % MOD;
		return *this;
	}
	C1097Int  operator-(const C1097Int& o)
	{
		return C1097Int((m_iData + MOD - o.m_iData) % MOD);
	}
	C1097Int  operator*(const C1097Int& o)const
	{
		return((long long)m_iData * o.m_iData) % MOD;
	}
	C1097Int& operator*=(const C1097Int& o)
	{
		m_iData = ((long long)m_iData * o.m_iData) % MOD;
		return *this;
	}
	bool operator<(const C1097Int& o)const
	{
		return m_iData < o.m_iData;
	}
	C1097Int pow(long long n)const
	{
		C1097Int iRet = 1, iCur = *this;
		while (n)
		{
			if (n & 1)
			{
				iRet *= iCur;
			}
			iCur *= iCur;
			n >>= 1;
		}
		return iRet;
	}
	C1097Int PowNegative1()const
	{
		return pow(MOD - 2);
	}
	int ToInt()const
	{
		return m_iData;
	}
private:
	int m_iData = 0;;
};

class Solution {
public:
	int countRoutes(vector<int>& locations, int start, int finish, int fuel) {
		const int n = locations.size();
		vector<vector<C1097Int<> >> dp(n, vector<C1097Int<>>(fuel + 1));
		dp[start][0] = 1;
		for (int f = 0; f < fuel; f++)
		{
			for (int l = 0; l < n; l++)
			{
				for (int l1 = 0; l1 < n; l1++)
				{
					if (l1 == l)
					{
						continue;
					}
					int f1 = f + abs(locations[l1] - locations[l]);
					if (f1 <= fuel)
					{
						dp[l1][f1] += dp[l][f];
					}
				}
			}
		}
		return std::accumulate(dp[finish].begin(), dp[finish].end(), C1097Int()).ToInt();
	}
};

测试用例

template<class T>
void Assert(const T& t1, const T& t2)
{
	assert(t1 == t2);
}

template<class T>
void Assert(const vector<T>& v1, const vector<T>& v2)
{
	if (v1.size() != v2.size())
	{
		assert(false);
		return;
	}
	for (int i = 0; i < v1.size(); i++)
	{
		Assert(v1[i], v2[i]);
	}

}

int main()
{	
	vector<int> locations;
	int start,  finish,  fuel;
	{
		Solution sln;
		locations = { 2, 3, 6, 8, 4 }, start = 1, finish = 3, fuel = 5;
		auto res = sln.countRoutes(locations, start, finish, fuel);
		Assert(4, res);
	}

	{
		Solution sln;
		locations = { 4, 3, 1 }, start = 1, finish = 0, fuel = 6;
		auto res = sln.countRoutes(locations, start, finish, fuel);
		Assert(5, res);
	}

	{
		Solution sln;
		locations = { 5, 2, 1 }, start = 0, finish = 2, fuel = 3;
		auto res = sln.countRoutes(locations, start, finish, fuel);
		Assert(0, res);
	}
}

小幅优化

排序后，向右（左）第一个油量不够的城市就停止。

class Solution {
public:
	int countRoutes(vector<int>& locations, int start, int finish, int fuel) {
		const int n = locations.size();
		const int startLoc = locations[start];
		const int finishLoc = locations[finish];
		sort(locations.begin(), locations.end());
		start = std::find(locations.begin(), locations.end(), startLoc)- locations.begin();
		finish = std::find(locations.begin(), locations.end(), finishLoc) - locations.begin();
		vector<vector<C1097Int<> >> dp(n, vector<C1097Int<>>(fuel + 1));
		dp[start][0] = 1;
		for (int f = 0; f < fuel; f++)
		{
			for (int l = 0; l < n; l++)
			{
				int use = 0;
				for (int l1 = l+1 ;(l1 < n )&&( f + (use = locations[l1] - locations[l]) <= fuel); l1++)
				{
					dp[l1][f+use] += dp[l][f];					
				}
				for (int l1 = l - 1; (l1 >= 0 ) && (f + (use = locations[l] - locations[l1]) <= fuel); l1--)
				{
					dp[l1][f + use] += dp[l][f];
				}
			}
		}
		return std::accumulate(dp[finish].begin(), dp[finish].end(), C1097Int()).ToInt();
	}
};

再次优化

填表顺序，还是从使用汽油少的到使用汽油多的。
前一个位置可能在当前位置的左边，也可能是右边。不失一般性，只讨论从左边过来。
假定当前使用汽油是f，位置是l。前一站使用的汽油是f1,位置是l1。则
行驶的路径等于消耗的汽油 $\to$ f-f1= l - l1 $\to$ f-l = f1-l1 性质一
符合性质一的状态分以下三类：
一，使用汽油比当前使用的汽油少。此状态就是本状态的前置状态。
二，使用汽油和当前状态使用的汽油一样，就是本状态。
三，使用汽油比当前汽油多，还没有更新。
结论： 符合性质一的状态路径和，就是本状态的路径和。
从右边来，类似：
f-f1=l1-l $\to$ f+l==f1+l1 性质二
m1，记录性质一；m2 记录性质二。
时间复杂度：将为O(nfuel)。

class Solution {
public:
	int countRoutes(vector<int>& locations, int start, int finish, int fuel) {
		const int n = locations.size();
		vector<vector<C1097Int<> >> dp(n, vector<C1097Int<>>(fuel + 1));
		unordered_map<int, C1097Int<>> m1,m2;
		dp[start][0] = 1;
		m1[0 - locations[start]] +=1 ;
		m2[0 + locations[start]] += 1;
		for (int f = 1; f <= fuel; f++)
		{
			for (int l = 0; l < n; l++)
			{
				dp[l][f] = m1[f- locations[l]]+ m2[f + locations[l]];
				m1[f - locations[l]] += dp[l][f];
				m2[f + locations[l]] += dp[l][f];
			}
		}
		return std::accumulate(dp[finish].begin(), dp[finish].end(), C1097Int()).ToInt();
	}
};

2023年2月第一版

class C1097Int
{
public:
C1097Int(int iData = 0) :m_iData(iData)
{
}
C1097Int operator+(const C1097Int& o)const
{
return C1097Int((m_iData + o.m_iData) % s_iMod);
}
C1097Int& operator+=(const C1097Int& o)
{
m_iData = (m_iData + o.m_iData) % s_iMod;
return this;
}
C1097Int operator(const C1097Int& o)const
{
return((long long)m_iData o.m_iData) % s_iMod;
}
C1097Int& operator=(const C1097Int& o)
{
m_iData =((long long)m_iData *o.m_iData) % s_iMod;
return *this;
}
int ToInt()const
{
return m_iData;
}
private:
int m_iData = 0;;
static const int s_iMod = 1000000007;
};

int operator+(int iData, const C1097Int& int1097)
{
int iRet = int1097.operator+(C1097Int(iData)).ToInt();
return iRet;
}

int& operator+=(int& iData, const C1097Int& int1097)
{
iData = int1097.operator+(C1097Int(iData)).ToInt();
return iData;
}

class Solution {
public:
int countRoutes(vector& locations, int start, int finish, int fuel) {
m_c = locations.size();
vector<vector> vFuelPos(fuel + 1, vector(m_c));
vFuelPos[fuel][start] = 1;
for (int iCurFuel = fuel-1; iCurFuel >= 0; iCurFuel–)
{
for (int iPos = 0; iPos < m_c; iPos++)
{
for (int iPrePos = 0; iPrePos < m_c; iPrePos++)
{
if (iPrePos == iPos)
{
continue;
}
const int iNeedFuel = iCurFuel + abs(locations[iPos] - locations[iPrePos]);
if (iNeedFuel <= fuel)
{
vFuelPos[iCurFuel][iPos] += vFuelPos[iNeedFuel][iPrePos];
}
}
}
}
C1097Int iNum = 0;
for (int iCurFuel = fuel ; iCurFuel >= 0; iCurFuel–)
{
iNum += vFuelPos[iCurFuel][finish];
}
return iNum.ToInt();
}
int m_c;
};

2023年9月版

class Solution {
public:
int countRoutes(vector& locations, int start, int finish, int fuel) {
m_c = locations.size();
int startValue = locations[start], finishValue = locations[finish];
std::sort(locations.begin(), locations.end());
start = std::lower_bound(locations.begin(), locations.end(), startValue) - locations.begin();
finish = std::lower_bound(locations.begin(), locations.end(), finishValue) - locations.begin();
m_vLeft.assign(m_c, vector<C1097Int<>>(fuel + 1));
m_vRight = m_vLeft;
int iRemain;
if ((start+1 < m_c)&&(( iRemain = fuel - locations[start+1]+ locations[start ]) >=0 ))
{
m_vRight[start+1][iRemain] = 1;
}
if ((start > 0) && ((iRemain = fuel - locations[start] + locations[start - 1]) >= 0))
{
m_vLeft[start-1][iRemain] = 1;
}
for (int iFuel = fuel-1; iFuel > 0; iFuel–)
{
for (int city = 0; city < m_c; city++)
{
RightMoveRight(iFuel, city, locations);
RightMoveLeft(iFuel, city, locations);
LeftMoveRight(iFuel, city, locations);
LeftMoveLeft(iFuel, city, locations);
}
}
C1097Int<> biRet = 0;
for (int iFuel = fuel; iFuel >= 0; iFuel–)
{
biRet += m_vLeft[finish][iFuel];
biRet += m_vRight[finish][iFuel];
}
if (start == finish)
{
biRet += 1;
}
return biRet.ToInt();
}
void RightMoveRight(int iFuel, int city, const vector& loc)
{
if (city + 1 >= m_c)
{
return;
}
const int iNeedFuel = loc[city + 1] - loc[city];
if (iNeedFuel > iFuel)
{
return;
}
m_vRight[city + 1][iFuel - iNeedFuel] += m_vRight[city][iFuel] * 2;
}
void RightMoveLeft(int iFuel, int city, const vector& loc)
{
if (0 == city)
{
return;
}
const int iNeedFuel = loc[city] - loc[city - 1];
if (iNeedFuel > iFuel)
{
return;
}
m_vLeft[city - 1][iFuel - iNeedFuel] += m_vRight[city][iFuel];
}
void LeftMoveRight(int iFuel, int city, const vector& loc)
{
if (city + 1 >= m_c)
{
return;
}
const int iNeedFuel = loc[city + 1] - loc[city];
if (iNeedFuel > iFuel)
{
return;
}
m_vRight[city + 1][iFuel - iNeedFuel] += m_vLeft[city][iFuel];
}
void LeftMoveLeft(int iFuel, int city, const vector& loc)
{
if (0 == city)
{
return;
}
const int iNeedFuel = loc[city] - loc[city - 1];
if (iNeedFuel > iFuel)
{
return;
}
m_vLeft[city - 1][iFuel - iNeedFuel] += m_vLeft[city][iFuel] * 2;
}
vector<vector<C1097Int<>>> m_vLeft, m_vRight;
int m_c;
};