思路:
这道题可以宽搜,深搜,最短路
代码:
#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
using namespace std;
queue<pair<int,int> > q;
int n,m,c[25],s;
long long f[25][1010],ans;
bool v[25][1010];
int Abs(int x)
{
return x >= 0 ? x : -x;
}
void bfs()
{
q.push(make_pair(s,1));
v[s][1] = 1; f[s][1] = 0;
while(!q.empty())
{
int x = q.front().first;
int y = q.front().second;
q.pop(); v[x][y] = 0;
for(int i = 1;i <= m;i ++)
{
int em = y + c[i];
if(em <= n && em >= 1)
{
if(f[i][em] > f[x][y] + Abs(i - x) + Abs(2 * c[i]))
{
f[i][em] = f[x][y] + Abs(i - x) + Abs(2 * c[i]);
if(!v[i][em]) q.push(make_pair(i,em)),v[i][em] = 1;
}
}
}
}
}
int main()
{
freopen("elevator.in","r",stdin);
freopen("elevator.out","w",stdout);
scanf("%d%d",&n,&m);
for(int i = 1;i <= m;i ++)
{
scanf("%d",&c[i]);
if(c[i] == 0) s = i;
}
memset(f,0x3f,sizeof(f));
ans = f[0][0];
bfs();
for(int i = 1;i <= m;i ++) ans = min(ans,f[i][n]);
if(ans != f[0][0]) printf("%d\n",ans);
else printf("-1\n");
fclose(stdin);
fclose(stdout);
return 0;
}
/*
6 3
-1 0 2
*/