例題8-13:環形跑道
題意:
環形跑道上的加油站有n個,第i個加油站加油pi單位,開到下個加油站需要qi單位,求一個起點使得從該點出發可以走完全程。
思路:
如果從l出發經過j最後在k加油站前面停下來,那麼從j出發一定也是到達不了k的,因爲從i出發可以途徑j,說明在路過j點還沒加油時有>=0的油,而從j點出發在還沒加油情況時有0油,所以可以用枚舉法枚舉起點,而上面情況i–k中間的點不必枚舉,模擬,複雜度爲O(n);
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<string>
#include<cmath>
#include<set>
#include<queue>
#include<map>
#include<stack>
#include<vector>
#include<list>
#include<deque>
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 10;
const double eps = 1e-6;
const int INF = 1 << 30;
int T, n, m;
int dis[maxn], val[maxn];
int judge(int x)
{
ll sum = 0, t;
for(int i = 0; i < n; i++)
{
if(i + x > n)t = i + x - n;
else t = i + x;
sum += val[t];
if(sum < dis[t])return i + x + 1;
sum -= dis[t];
}
return x;
}
int main()
{
scanf("%d",&T);
int cases = 0;
while(T--)
{
scanf("%d",&n);
for(int i = 1; i <= n; i++)scanf("%d",&val[i]);
for(int i = 1; i <= n; i++)scanf("%d",&dis[i]);
int flag = 0, ans;
for(int i = 1; i <= n; i++)
{
if(judge(i) == i){flag = 1; ans = i; break;}
else
{
i = judge(i) - 1;
if(i > n)break;
}
}
printf("Case %d: ",++cases);
if(flag)
{
printf("Possible from station %d\n",ans);
}
else printf("Not possible\n");
}
return 0;
}