POJ 3282 Ferry Loading IV(简单模拟)
2014-08-14 20:17
363 查看
Description Before bridges were common, ferries were used to transport cars across rivers. River ferries, unlike their larger cousins, run on a guide line and are powered by the river's current. Cars drive onto the ferry from one end, the ferry crosses the river, and the cars exit from the other end of the ferry. There is an l-meter-long ferry that crosses the river. A car may arrive at either river bank to be transported by the ferry to the opposite bank. The ferry travels continuously back and forth between the banks so long as it is carrying a car or there is at least one car waiting at either bank. Whenever the ferry arrives at one of the banks, it unloads its cargo and loads up cars that are waiting to cross as long as they fit on its deck. The cars are loaded in the order of their arrival; ferry's deck accommodates only one lane of cars. The ferry is initially on the left bank where it broke and it took quite some time to fix it. In the meantime, lines of cars formed on both banks that await to cross the river. Input The first line of input contains c, the number of test cases. Each test case begins with l, m. m lines follow describing the cars that arrive in this order to be transported. Each line gives the length of a car (in centimeters), and the bank at which the car arrives ("left" or "right"). Output For each test case, output one line giving the number of times the ferry has to cross the river in order to serve all waiting cars. Sample Input 4 20 4 380 left 720 left 1340 right 1040 left 15 4 380 left 720 left 1340 right 1040 left 15 4 380 left 720 left 1340 left 1040 left 15 4 380 right 720 right 1340 right 1040 right Sample Output 3 3 5 6 简单模拟,最后20分钟开的这道题,写的急了,wa了3发,真想骂自己SB。 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<limits.h> #include<queue> using namespace std; int t,len,n; int main() { char str[15]; int x; scanf("%d",&t); while(t--) { queue<int>l; queue<int>r; scanf("%d%d",&len,&n); for(int i=0;i<n;i++) { scanf("%d%s",&x,str); if(str[0]=='l') l.push(x); else r.push(x); } int d=1;//记录船在哪边1代表左 int num=0; while(!l.empty()||!r.empty()) { int sum=len*100; if(d==1) { int s=0; while(!l.empty()&&s+l.front()<=sum) { s+=l.front(); l.pop(); } d=-d; num++; } else { int s=0; while(!r.empty()&&s+r.front()<=sum) { s+=r.front(); r.pop(); } d=-d; num++; } } printf("%d\n",num); } return 0; } |
相关文章推荐
- POJ 3282 Ferry Loading IV(模拟)
- POJ 3282 Ferry Loading IV(模拟,队列)
- POJ 3282 Ferry Loading IV(模拟,队列)
- POJ 3282 Ferry Loading IV 可能会
- POJ-2336 Ferry Loading II(简单DP)
- POJ-2336 Ferry Loading II(简单DP)
- (简单模拟——筛选法模拟2.2.1)POJ 1316 Self Numbers(生成数、自数)
- poj 1573 简单模拟
- POJ1318 Word Amalgamation 简单模拟
- POJ 1573 Robot Motion(简单模拟)
- (简单模拟2.4.2)POJ 1207 The 3n + 1 problem(直叙式模拟——输入时前一个数可能比后一个数要大,但输出时原样输出)
- poj 3207 Ikki's Story IV - Panda's Trick(2-sat简单应用)
- poj 1552 Doubles(简单的模拟一下)
- POJ 3652 Persistent Bits 简单字符串模拟
- Parencodings(poj1068简单模拟)
- (简单模拟2.1.1)POJ 2017 Speed Limit(跨项累加sumLen += (t2-t1)*v )
- POJ 2609 Ferry Loading 双塔DP
- POJ1250 Tanning Salon 简单模拟
- POJ 3652 Persistent Bits 简单字符串模拟
- POJ 1573 && hdu 1035(简单模拟)