usaco2018 feb cu Taming the Herd

usaco2018 feb cu Taming the Herd

​​ Early in the morning, Farmer John woke up to the sound of splintering wood. It was the cows, and they were breaking out of the barn again!

Farmer John was sick and tired of the cows’ morning breakouts, and he decided enough was enough: it was time to get tough. He nailed to the barn wall a counter tracking the number of days since the last breakout. So if a breakout occurred in the morning, the counter would be 0

0

that day; if the most recent breakout was 3

3

days ago, the counter would read 3

3

. Farmer John meticulously logged the counter every day.

The end of the year has come, and Farmer John is ready to do some accounting. The cows will pay, he says! But lo and behold, some entries of his log are missing!

Farmer John is confident that the he started his log on the day of a breakout. Please help him determine, out of all sequences of events consistent with the log entries that remain, the minimum and maximum number of breakouts that may have take place over the course of the logged time.

INPUT FORMAT (file taming.in):

The first line contains a single integer N

N (1≤N≤100

1≤N≤100 ), denoting the number of days since Farmer John started logging the cow breakout counter.The second line contains N

N space-separated integers. The i

i th integer is either −1

−1 , indicating that the log entry for day i

i is missing, or a non-negative integer ai

ai (at most 100

100 ), indicating that on day i

i the counter was at ai

ai .

OUTPUT FORMAT (file taming.out):

If there is no sequence of events consistent with Farmer John’s partial log and his knowledge that the cows definitely broke out of the barn on the morning of day 1

1 , output a single integer −1

−1 . Otherwise, output two space-separated integers m

m followed by M

M , where m

m is the minimum number of breakouts of any consistent sequence of events, and M

M is the maximum.

SAMPLE INPUT:

4 -1 -1 -1 1 SAMPLE OUTPUT:

2 3 In this example, we can deduce that a breakout had to occur on day 3. Knowing that a breakout also occurred on day 1, the only remaining bit of uncertainty is whether a breakout occurred on day 2. Hence, there were between 2 and 3 breakouts in total.

Problem credits: Dhruv Rohatgi

-1代表当天数据丢失 如果牛出逃则计数器变成0 否则计数器显示的是距离上一次牛出逃

现在给一个计数器的序列 求牛最多和最少出逃几次

贪心 中间有-1 全部都认为出逃或者都不出逃 注意好边界条件即可

#include#include#include#define N 110using namespace std;inline char gc(){ static char now[1<<16],*S,*T; if (T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;} return *S++;}inline int read(){ int x=0,f=1;char ch=gc(); while(ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=gc();} while(ch<='9'&&ch>='0') x=x*10+ch-'0',ch=gc(); return x*f;}int ans1;//-1's numint ans2,n,x[N];// mininum timesbool visit[N];int main(){ freopen("taming.in","r",stdin); freopen("taming.out","w",stdout); n=read();bool flag=0;ans2=1; for (int i=1;i<=n;++i) x[i]=read(); for (int i=n;i;--i){ if (visit[i]) continue; if (x[i]==-1){if (i!=1)++ans1;continue;} int now=x[i];++ans2; for (int j=0;j<=x[i];++j,--now){ if (i-j<0) {flag=1;break;} visit[i-j]=1; if (x[i-j]!=now&&x[i-j]!=-1) {flag=1;break;} if (i-j==1) --ans2; } } if (flag) {puts("-1");return 0;} printf("%d %d",ans2,ans2+ans1); return 0;}

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