zoukankan      html  css  js  c++  java
  • POJ 1724 ROADS

    ROADS
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 9118   Accepted: 3383

    Description

    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.

    Input

    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.

    Output

    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.

    Sample Input

    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
    

    Sample Output

    11

    Source

    CEOI 1998
      用spfa即可
    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <queue>
    #define N1 101
    #define N2 10001
    #define INF 0x7ffffff
    using namespace std;
    bool inque[N1][N2];
    int sum[N1][N2];
    class num
    {
        public:
        int id,take;
        num(int l,int r):id(l),take(r){}
    };
    struct link
    {
        int e,take,len,next;
    }a[N2];
    int b[N1],k,n,m,Top;
    int main()
    {
        //freopen("data1.in","r",stdin);
        void addeage(int s,int d,int l,int t);
        int spfa();
        while(scanf("%d %d %d",&k,&n,&m)!=EOF)
        {
            Top = 0;
            memset(b,-1,sizeof(b));
            for(int i=0;i<=m-1;i++)
            {
                int s,d,l,t;
                scanf("%d %d %d %d",&s,&d,&l,&t);
                addeage(s,d,l,t);
            }
            int t=spfa();
            printf("%d
    ",t);
        }
        return 0;
    }
    void addeage(int s,int d,int l,int t)
    {
        a[Top].e = d;
        a[Top].len = l;
        a[Top].take = t;
        a[Top].next = b[s];
        b[s] = Top;
        Top++;
    }
    int spfa()
    {
        memset(inque,false,sizeof(inque));
        for(int i = 1;i <= n; i++)
        {
            for(int j=0; j <= k; j++)
            {
                sum[i][j] = INF;
            }
        }
        queue<num> que;
        que.push(num(1,0));
        inque[1][0] = true;
        sum[1][0] = 0;
        while(!que.empty())
        {
            num tag =  que.front();
            int id1 = tag.id;
            int take1 = tag.take;
            que.pop();
            inque[id1][take1] = false;
            for(int i = b[id1]; i!=-1; i=a[i].next)
            {
                int id2 = a[i].e;
                int len2 = a[i].len;
                int take2 = a[i].take;
                if(take1+take2<=k&&sum[id2][take1+take2]>sum[id1][take1]+len2)
                {
                    sum[id2][take1+take2]=sum[id1][take1]+len2;
                    if(!inque[id2][take1+take2])
                    {
                        que.push(num(id2,take1+take2));
                        inque[id2][take1+take2] = true;
                    }
                }
            }
        }
        int Min = INF;
        for(int i=0;i<=k;i++)
        {
            Min = min(sum[n][i],Min);
        }
        if(Min==INF)
        {
            return -1;
        }
        return Min;
    }
    


  • 相关阅读:
    AngularJS Directive 隔离 Scope 数据交互
    Web(Jsp+ Servlet)开发中如何解决中文乱码问题
    MySQL中进行模糊搜索的一些问题
    RequireJS 模块的定义与加载
    Mysql 正则表达式 判断字段值不包含数字
    使用命令行将Excel数据表导入Mysql中的方法小结
    js 去除字符串左右两端的空格
    js 计算两个时间差
    在MySQL中创建实现自增的序列(Sequence)的教程
    ajax post传值
  • 原文地址:https://www.cnblogs.com/javawebsoa/p/3209227.html
Copyright © 2011-2022 走看看