題目鏈接：

Ferry Loading II

Time Limit:?1000MS	?	Memory Limit:?65536K
Total Submissions:?3946	?	Accepted:?1985

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 a ferry across the river that can take n cars across the river in t minutes and return in t minutes. m cars arrive at the ferry terminal by a given schedule. What is the earliest time that all the cars can be transported across the river? What is the minimum number of trips that the operator must make to deliver all cars by that time?

Input

The first line of input contains c, the number of test cases. Each test case begins with n, t, m. m lines follow, each giving the arrival time for a car (in minutes since the beginning of the day). The operator can run the ferry whenever he or she wishes, but can take only the cars that have arrived up to that time.

Output

For each test case, output a single line with two integers: the time, in minutes since the beginning of the day, when the last car is delivered to the other side of the river, and the minimum number of trips made by the ferry to carry the cars within that time.?

You may assume that 0 < n, t, m < 1440. The arrival times for each test case are in non-decreasing order.

Sample Input

2 2 10 10 0 10 20 30 40 50 60 70 80 90 2 10 3 10 30 40

Sample Output

100 5 50 2

題意：

給m輛車的到達岸邊的時間,現(xiàn)在給你一個輪渡能運車的數(shù)量,和單程的時間,現(xiàn)在問把這些車運過去的最短時間是多少,在這個時間中的 最少運送次數(shù)是多少？

思路：

dp[i]表示運送前i個要用的時間,num[i]表示在dp[i]的時間內(nèi)的最少次數(shù);相鄰的車在一塊運,轉(zhuǎn)移方程看代碼吧;

AC代碼：

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> //#include <bits/stdc++.h> #include <stack> #include <map>using namespace std;#define For(i,j,n) for(int i=j;i<=n;i++) #define mst(ss,b) memset(ss,b,sizeof(ss));typedef long long LL;template<class T> void read(T&num) {char CH; bool F=false;for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());F && (num=-num); } int stk[70], tp; template<class T> inline void print(T p) {if(!p) { puts("0"); return; }while(p) stk[++ tp] = p%10, p/=10;while(tp) putchar(stk[tp--] + '0');putchar('\n'); }const LL mod=1e9+7; const double PI=acos(-1.0); const int inf=1e9; const int N=1e5+10; const int maxn=2e3+14; const double eps=1e-12;int a[maxn],dp[maxn],num[maxn];int main() {int T;read(T);while(T--){int n,m,t;read(n);read(t);read(m);For(i,1,m)read(a[i]);For(i,1,m)dp[i]=inf,num[i]=0;dp[0]=0;num[0]=0;For(i,1,m){for(int j=max(0,i-n);j<i;j++){if(j==0){dp[i]=a[i]+t,num[i]=1;continue;}if(dp[i]>max(dp[j]+t,a[i])+t)dp[i]=max(dp[j]+t,a[i])+t,num[i]=num[j]+1;else if(dp[i]==max(dp[j]+t,a[i])+t)num[i]=min(num[i],num[j]+1);}}cout<<dp[m]<<" "<<num[m]<<"\n";}return 0; }

　　

轉(zhuǎn)載于:https://www.cnblogs.com/zhangchengc919/p/5754980.html

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