原题网址：https://open.kattis.com/problems/driver
Crazy Driver
In the Linear City, there are N gates arranged in a straight line. The gates are labelled from 1 to N. Between adjacent gates, there is a bidirectional road. Each road takes one hour to travel and has a toll fee. Since the roads are narrow, you can only travel from gates to gates but cannot U-turn between gates.
Crazy driver Gary starts at Gate 1 at time 0 and he wants to drive through Gate N while minimizing the cost of travelling. However, Gate i only allows a car to pass through after a certain time Ti. As Gary is crazy, his car will always be traveling on any one of the roads, i.e., it will not stop at a gate. What is the minimum cost for him to drive through Gate N ?
As an example, consider the sample input below. An optimal solution is the following:
- Gate 1 to Gate 2 (cost 5)
- Gate 2 to Gate 1 (cost 5)
- Gate 1 to Gate 2 to Gate 3 (cost 9)
- Go between Gate 3 and Gate 4 until 7-th hour (cost 6)
- Go to and pass through Gate 5(cost 8)
Input
The first line contains an integer, N(2≤N≤105), the number of gates. The second line has N−1 integers, C1,…,CN−1. Ci (1≤Ci≤106) represents the toll fee of the road between Gate i and Gate i+1. The third line has N integers, T1,…,TN. Ti (0≤Ti≤106) represents the opening time (in hour) for each gate. T1 will always be 0.
Output
Output an integer representing the minimum cost of traveling.
|
Sample Input 1 |
Sample Output 1 |
|
5 5 4 2 8 0 2 4 4 8 |
33 |
题意：n个门编号1~n,从门i到i+1有一条双向通路，每条路花费的时间都是1小时，每条路花的路费分别是Ci, 每个门开的时刻分别是Ti，一个司机从门1开到门n，中间不停车，即如果到达门i的时候门没开就必须往返于前面的路上直到门开的时刻，问到门n最少花多少路费。
记录每扇门之前的路的最小路费。
#include <algorithm> #include <cstring> #include <string.h> #include <iostream> #include <list> #include <map> #include <set> #include <stack> #include <string> #include <utility> #include <vector> #include <cstdio> #include <cmath> #define LL long long #define N 100005 #define INF 0x3ffffff using namespace std; int n; int c[N]; //门i-1到门i的路费是Ci int m[N]; //门i之前的路的路费最小值 int t[N]; //每个门开的时刻 int main() { scanf("%d",&n); for(int i=1;i<=n-1;i++) { scanf("%d",&c[i]); if(i==1) m[i]=c[i]; else m[i]=min(m[i-1],c[i]); } for(int i=0;i<n;i++) { scanf("%d",&t[i]); } int tt=0; //当前时刻 int i=0; long long ret=0; while(i<n) { i++; tt++; ret+=(long long)(c[i]); int tmp=t[i]-tt; //离门开还有多久 while(tmp>0){ tmp-=2; ret+=(long long)(m[i]*2); tt+=2; } } cout<<ret<<endl; return 0; }