cf_288 C

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题目链接:http://codeforces.com/contest/508/problem/C

思路:在第一个鬼魂来之前先点燃所需的r根蜡烛,并把这r根蜡烛的点燃时间放入一个队列中,那么在第二个来的时候,我们先看之前最先点燃的蜡烛有没有灭,如果灭了,那么得重新点一根,并把点燃时间放入队列中,一直循环到所有ghost都访问结束,此时输出所需的蜡烛的根数。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<map>#include<set>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;#define ll __int64int n,m,k;#define N  100010int a[310];queue<int> q;int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);//    freopen("out.txt","w",stdout);#endif    int t,r;    while(scanf("%d%d%d",&m,&t,&r)!=EOF){    while(!q.empty()){    q.pop();    }    for(int i=0;i<m;i++)    sf(a[i]);    if(t<r){//如果蜡烛持续燃烧的时间小于它最小数量,那么根本达不到所需要求(because点燃蜡烛需要1s)    printf("-1\n");    continue;    }    int ans=0;    for(int i=r;i>=1;i--)//迎接第一个之前点燃r个蜡烛,点燃时间为a[0]-i    q.push(a[0]-i);    ans += r;    for(int i=1;i<m;i++){    int ct=0;    while(!q.empty() && q.front()+t<a[i]){//队列不为空并且最先点燃的蜡烛已经熄灭了,那么得再点燃一根    q.pop();    ct++;    }    ans += ct;    for(int j=ct;j>=1;j--)//把点燃的蜡烛放进队列    q.push(a[i]-j);    }    printf("%d\n",ans);   }return 0;}


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