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.

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 0x3ffffffusing namespace std;int n;int c[N];               //门i-1到门i的路费是Ciint 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;}

2016 acm香港网络赛 F题. Crazy Driver(水题)