C++ 具有加油站任务变化的DP算法
我有一个问题,我从a点旅行到B点,距离定义为C++ 具有加油站任务变化的DP算法,c++,algorithm,data-structures,dynamic-programming,greedy,C++,Algorithm,Data Structures,Dynamic Programming,Greedy,我有一个问题,我从a点旅行到B点,距离定义为l。在我的车里,我有一个油箱,它只能让我跑b公里。在n的地方有加油站,加油的费用(我只能完全加油)。我从满油箱开始 从高速公路的起点到终点,成本最低的最有效的算法是什么 我最近的想法: 使用窗口滑动最小算法查找(长度为b)成本最低的站点。代码: int n, range, targetDist; scanf("%d %d %d", &n, &targetDist, &range); int di, ci; for (int k
l
。在我的车里,我有一个油箱,它只能让我跑b
公里。在n
的地方有加油站,加油的费用(我只能完全加油)。我从满油箱开始
从高速公路的起点到终点,成本最低的最有效的算法是什么
我最近的想法:
使用窗口滑动最小算法查找(长度为b
)成本最低的站点。代码:
int n, range, targetDist;
scanf("%d %d %d", &n, &targetDist, &range);
int di, ci;
for (int k = 1; k <= n; k++)
{
scanf("%d %d", &di, &ci);
D[k] = di;
C[k] = ci;
}
D[0] = C[0] = bestcost[0] = 0;
for (int i = 1; i < n+1; ++i)
{
bestcost[i] = 1 << 30;
for (int j = 0; j < i; ++j)
{
int xx = bestcost[j] + C[i];
if (D[i] - D[j] <= range && xx < bestcost[i])
{
bestcost[i] = xx;
printf("(%d, %d)\n", i, bestcost[i]);
}
}
}
输出为:
(1, 1)
(2, 2)
(3, 4)
所以它对应于成本(i,成本(i))-要到达第i站,我必须支付成本(i)
如何利用这些信息找到整个距离的最小成本?我找到了一种在O(n)时间内使用滑动窗口算法的方法:
#include <vector>
#include <algorithm>
#include <deque>
#include <stdio.h>
typedef unsigned long long int ulli;
using namespace std;
void sliding_window_minimum(vector<pair<ulli, ulli>> st, ulli n, ulli targetDist, ulli range)
{
deque<pair<ulli, ulli>> minimize;
ulli j = 0;
for (ulli i = 0; i < n; i++)
{
if (st[i].first <= range)
{
while (!(minimize.empty()) && (minimize.back().second > st[i].second))
{
minimize.pop_back();
}
minimize.push_back(st[i]);
j++;
}
else
{
break;
}
}
for (ulli k = j; k < n; k++)
{
while (!(minimize.empty()) && ((st[k].first - minimize.front().first) > range))
{
minimize.pop_front();
}
if (minimize.empty()) {
break;
}
ulli tempCost = st[k].second + minimize.front().second;
while (!(minimize.empty()) && (minimize.back().second > tempCost))
{
minimize.pop_back();
}
minimize.push_back(make_pair(st[k].first, tempCost));
}
while (!(minimize.empty()) && ((targetDist - minimize.front().first) > range))
{
minimize.pop_front();
}
if (minimize.empty())
{
printf("NIE\n");
}
else
{
printf("%llu", minimize.front().second);
}
}
int main()
{
ulli n, d, b;
scanf("%llu %llu %llu", &n, &d, &b);
if (b >= d)
{
printf("%d", 0);
return 0;
}
int temp = n;
ulli d1, c1;
vector<pair<ulli, ulli>> st;
st.reserve(n+1);
while (temp--)
{
scanf("%llu %llu", &d1, &c1);
st.push_back(make_pair(d1, c1));
}
sliding_window_minimum(st, n, d, b);
}
#包括
#包括
#包括
#包括
typedef无符号长整型ulli;
使用名称空间std;
无效滑动窗口最小值(矢量st、ulli n、ulli targetDist、ulli范围)
{
德克最小化;
ulli j=0;
对于(ulli i=0;i范围)
{
最小化.pop_front();
}
if(最小化.empty()){
打破
}
ulli tempCost=st[k].second+minimize.front().second;
而(!(minimize.empty())&&(minimize.back().second>tempCost))
{
最小化.pop_back();
}
最小化。向后推(生成一对(st[k]。首先,tempCost));
}
而(!(minimize.empty())&((targetDist-minimize.front().first)>范围)
{
最小化.pop_front();
}
if(最小化.empty())
{
printf(“NIE\n”);
}
其他的
{
printf(“%llu”,minimize.front().second);
}
}
int main()
{
乌利n,d,b;
scanf(“%llu%llu%llu”、&n、&d、&b);
如果(b>=d)
{
printf(“%d”,0);
返回0;
}
内部温度=n;
ulli d1,c1;
向量st;
圣保留地(n+1);
while(临时--)
{
scanf(“%llu%llu”、&d1和&c1);
st.push_back(形成一对(d1,c1));
}
滑动窗口最小值(st、n、d、b);
}
显然,问题已经讨论过了。
#include <vector>
#include <algorithm>
#include <deque>
#include <stdio.h>
typedef unsigned long long int ulli;
using namespace std;
void sliding_window_minimum(vector<pair<ulli, ulli>> st, ulli n, ulli targetDist, ulli range)
{
deque<pair<ulli, ulli>> minimize;
ulli j = 0;
for (ulli i = 0; i < n; i++)
{
if (st[i].first <= range)
{
while (!(minimize.empty()) && (minimize.back().second > st[i].second))
{
minimize.pop_back();
}
minimize.push_back(st[i]);
j++;
}
else
{
break;
}
}
for (ulli k = j; k < n; k++)
{
while (!(minimize.empty()) && ((st[k].first - minimize.front().first) > range))
{
minimize.pop_front();
}
if (minimize.empty()) {
break;
}
ulli tempCost = st[k].second + minimize.front().second;
while (!(minimize.empty()) && (minimize.back().second > tempCost))
{
minimize.pop_back();
}
minimize.push_back(make_pair(st[k].first, tempCost));
}
while (!(minimize.empty()) && ((targetDist - minimize.front().first) > range))
{
minimize.pop_front();
}
if (minimize.empty())
{
printf("NIE\n");
}
else
{
printf("%llu", minimize.front().second);
}
}
int main()
{
ulli n, d, b;
scanf("%llu %llu %llu", &n, &d, &b);
if (b >= d)
{
printf("%d", 0);
return 0;
}
int temp = n;
ulli d1, c1;
vector<pair<ulli, ulli>> st;
st.reserve(n+1);
while (temp--)
{
scanf("%llu %llu", &d1, &c1);
st.push_back(make_pair(d1, c1));
}
sliding_window_minimum(st, n, d, b);
}