2016 acm香港网络赛 F题. Crazy Driver(水题)

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 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;
}