2017 google code jam D轮

题目描述：
When you travel, you like to spend time sightseeing in as many cities as possible, but sometimes you might not be able to because you need to catch the bus to the next city. To maximize your travel enjoyment, you decide to write a program to optimize your schedule.
You begin at city 1 at time 0 and plan to travel to cities 2 to N in ascending order, visiting every city. There is a bus service from every city i to the next city i + 1. The i-th bus service runs on a schedule that is specified by 3 integers: Si, Fi and Di, the start time, frequency and ride duration. Formally, this means that there is a bus leaving from city i at all times Si+ xFi, where x is an integer and x ≥ 0, and the bus takes Di time to reach city i + 1.
At each city between 1 and N - 1, inclusive, you can decide to spend Ts time sightseeing before waiting for the next bus, or you can immediately wait for the next bus. You cannot go sightseeing multiple times in the same city. You may assume that boarding and leaving buses takes no time. You must arrive at city N by time Tf at the latest. (Note that you cannot go sightseeing in city N, even if you arrive early. There’s nothing to see there!)
What is the maximum number of cities you can go sightseeing in?
Input

The input starts with one line containing one integer T, which is the number of test cases. T test cases follow.
Each test case begins with a line containing 3 integers, N, Ts and Tf, representing the number of cities, the time taken for sightseeing in any city, and the latest time you can arrive in city N.
This is followed by N - 1 lines. On the i-th line, there are 3 integers, Si, Fi and Di, indicating the start time, frequency, and duration of buses travelling from city i to city i + 1.
Output

For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the maximum number of cities you can go sightseeing in such that you can still arrive at city N by time Tf at the latest. If it is impossible to arrive at city N by time Tf, output Case #x: IMPOSSIBLE.
Limits

1 ≤ T ≤ 100.
Small dataset

2 ≤ N ≤ 16.
1 ≤ Si ≤ 5000.
1 ≤ Fi ≤ 5000.
1 ≤ Di ≤ 5000.
1 ≤ Ts ≤ 5000.
1 ≤ Tf ≤ 5000.
Large dataset

2 ≤ N ≤ 2000.
1 ≤ Si ≤ 109.
1 ≤ Fi ≤ 109.
1 ≤ Di ≤ 109.
1 ≤ Ts ≤ 109.
1 ≤ Tf ≤ 109.
Sample

Input

Output

4
4 3 12
3 2 1
6 2 2
1 3 2
3 2 30
1 2 27
3 2 1
4 1 11
2 1 2
4 1 5
8 2 2
5 10 5000
14 27 31
27 11 44
30 8 20
2000 4000 3

Case #1: 2
Case #2: 0
Case #3: IMPOSSIBLE
Case #4: 4

我的解题思路，主要是一个回溯的思想，每一次都会选择是否要参观当前城市，如果参观时间是否足够，以下是我的源码。

#include <iostream>#include <vector>#include <fstream>using namespace std;int find(vector<vector<double> > nums, double ts, int count);bool judge(vector<vector<double> > nums, double ts);int main(){    ifstream ifile;    ifile.open("in.txt");    ofstream ofile;               //定义输出文件    ofile.open("out.txt");     //作为输出文件打开    int T;    ifile>>T;    vector<vector<double> > nums;    for(int i=0; i<T; i++){        int N, Ts, Tf;        ifile>>N;        ifile>>Ts;        ifile>>Tf;        vector<double> num;        num.push_back(N);        num.push_back(Ts);        num.push_back(Tf);        nums.push_back(num);        for(int j=0; j<N-1; j++){            int Si, Fi, Di;            vector<double> mid;            ifile>>Si;            ifile>>Fi;            ifile>>Di;            mid.push_back(Si);            mid.push_back(Fi);            mid.push_back(Di);            nums.push_back(mid);        }    }    ifile.close();    int j=0, count = 0;    for(int i=j; j<nums.size(); ){        i = j;        count++;        vector<vector<double> > child;        child.push_back(nums[i]);        for(j=i+1; j< i+nums[i][0];j++){            child.push_back(nums[j]);        }        if(!judge(child, 0)){            cout<<"Case #"<<count<<": IMPOSSIBLE"<<endl;            ofile<<"Case #"<<count<<": IMPOSSIBLE"<<endl;        }        else{            int res = find(child, 0, 0);            if(res < 0){                cout<<"Case #"<<count<<": IMPOSSIBLE"<<endl;                ofile<<"Case #"<<count<<": IMPOSSIBLE"<<endl;            }            else{                cout<<"Case #"<<count<<": "<<res<<endl;                ofile<<"Case #"<<count<<": "<<res<<endl;            }        }    }    ofile.close();}int find(vector<vector<double> > nums, double ts, int count){    count ++;    if(nums.size()==0){        return 0;    }    double N, Ts, Tf;    N = nums[0][0];    Ts = nums[0][1];    Tf = nums[0][2];    double Si, Fi, Di;    if(N==1){        return 0;    }    else{        Si = nums[1][0];        Fi = nums[1][1];        Di = nums[1][2];        nums.erase(nums.begin()+1);        double newts;        while(ts > Si){            Si += Fi;        }        newts = Si + Di;        double newts2;        if(ts+Ts <= Si){            newts2 = Si + Di;        }        else{            while(ts + Ts > Si){                Si += Fi;            }            newts2 = Si+Di;        }        nums[0][0] -= 1;        //判断是否能够在下余生        if(judge(nums, newts2)){            return max(1+find(nums, newts2, count), find(nums, newts, count));        }        else if(judge(nums, newts)){            return find(nums, newts, count);        }        else{            return -100;        }    }}bool judge(vector<vector<double> > nums, double ts){    double Tf = nums[0][2];    if(nums.size()==1){        if(ts <= Tf){            return true;        }        else{            return false;        }    }    for(int i=1; i< nums.size(); i++){        double Si = nums[i][0];        double Fi = nums[i][1];        double Di = nums[i][2];        if(Si > Tf || Di > Tf || Si+Di>Tf){            return false;        }        if(ts < Si){            ts = Si+Di;        }        else{            while(ts > Si){                Si += Fi;            }            ts = Si+Di;        }        if(ts > Tf){            return false;        }    }    return true;}

源码通过第一题的小数据集测试，但是大数据集没有通过，大家一起来看看呀，欢迎指正。大数据集如下：