ROADS

ROADS

总时间限制:
1000ms
内存限制:
65536kB

描述
N cities named with numbers 1 … N are connected with one-way roads. Each road has two parameters associated with it : the road length and the toll that needs to be paid for the road (expressed in the number of coins).
Bob and Alice used to live in the city 1. After noticing that Alice was cheating in the card game they liked to play, Bob broke up with her and decided to move away - to the city N. He wants to get there as quickly as possible, but he is short on cash.

We want to help Bob to find the shortest path from the city 1 to the city N that he can afford with the amount of money he has.

输入
The first line of the input contains the integer K, 0 <= K <= 10000, maximum number of coins that Bob can spend on his way.
The second line contains the integer N, 2 <= N <= 100, the total number of cities.

The third line contains the integer R, 1 <= R <= 10000, the total number of roads.

Each of the following R lines describes one road by specifying integers S, D, L and T separated by single blank characters :

    S is the source city, 1 <= S <= N
    D is the destination city, 1 <= D <= N
    L is the road length, 1 <= L <= 100
    T is the toll (expressed in the number of coins), 0 <= T <=100


Notice that different roads may have the same source and destination cities. 

输出
The first and the only line of the output should contain the total length of the shortest path from the city 1 to the city N whose total toll is less than or equal K coins.
If such path does not exist, only number -1 should be written to the output.
样例输入

5
6
7
1 2 2 3
2 4 3 3
3 4 2 4
1 3 4 1
4 6 2 1
3 5 2 0
5 4 3 2

样例输出

11

关于这条题目，我后面有空会把视频补上

//by Gary
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
//使用邻接表
int K,N,R;//钱数，城市数，道路数
struct Road
{
    int d,L,t;//终点，长度，过路费
};
vector<vector<Road>> G(150);
int midL[150][10100];
int minLen;
int totalLen;
int totalCost;
bool visited[150];
void dfs(int s)//从s出发
{
    if(s==N)
    {//起电=终点
        minLen=min(minLen,totalLen);
        return ;
    }
    for(int i=0;i<G[s].size();++i)
    {
        Road r=G[s][i];
        //看看钱够不够
        if(totalCost+r.t >K)//可行性剪枝
            continue;

        if(!visited[r.d])//没走过
        {
            if(totalLen+r.L>=minLen)//最优性剪枝
                continue;
            if(midL[r.d][totalCost+r.t]<=totalLen+r.L)
                continue;
            midL[r.d][totalCost+r.t]=totalLen+r.L;
            totalLen+=r.L;
            totalCost+=r.t;
            visited[r.d]=1;
            dfs(r.d);
            //撤销
            visited[r.d]=0;
            totalCost-=r.t;
            totalLen-=r.L;
        }
    }
}

int main()
{
    cin>>K>>N>>R;
    for(int i=0;i<R;++i)
    {
        int s;
        Road r;
        cin>>s>>r.d>>r.L>>r.t;
        if(s!=r.d)
        G[s].push_back(r);//加到邻接表里面
    }
    memset(visited,0,sizeof(visited));
    totalLen=0;
    totalCost=0;
    minLen=1<<30;
    visited[1]=1;
        for(int i=0;i<150;++i)
        for(int j=0;j<10100;++j)
        midL[i][j]=1<<30;
    dfs(1);
    for(int i=0;i<150;++i)
        for(int j=0;j<10100;++j)
        midL[i][j]=1<<30;
    if(minLen<(1<<30))
    {
        cout<<minLen<<endl;
    }

    else cout<<"-1"<<endl;
    return 0;
}

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