poj-2336 Ferry Loading II(dp)
题目链接:
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 3946 | Accepted: 1985 |
Description
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
Output
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 题意: 给m辆车的到达岸边的时间,现在给你一个轮渡能运车的数量,和单程的时间,现在问把这些车运过去的最短时间是多少,在这个时间中的 最少运送次数是多少? 思路: dp[i]表示运送前i个要用的时间,num[i]表示在dp[i]的时间内的最少次数;相邻的车在一块运,转移方程看代码吧; AC代码:
#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;
}
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