zoukankan      html  css  js  c++  java
  • poj 3159 Candies 差分约束

    Candies
    Time Limit: 1500MS   Memory Limit: 131072K
    Total Submissions: 22177   Accepted: 5936

    Description

    During the kindergarten days, flymouse was the monitor of his class. Occasionally the head-teacher brought the kids of flymouse’s class a large bag of candies and had flymouse distribute them. All the kids loved candies very much and often compared the numbers of candies they got with others. A kid A could had the idea that though it might be the case that another kid B was better than him in some aspect and therefore had a reason for deserving more candies than he did, he should never get a certain number of candies fewer than B did no matter how many candies he actually got, otherwise he would feel dissatisfied and go to the head-teacher to complain about flymouse’s biased distribution.

    snoopy shared class with flymouse at that time. flymouse always compared the number of his candies with that of snoopy’s. He wanted to make the difference between the numbers as large as possible while keeping every kid satisfied. Now he had just got another bag of candies from the head-teacher, what was the largest difference he could make out of it?

    Input

    The input contains a single test cases. The test cases starts with a line with two integers N and M not exceeding 30 000 and 150 000 respectively. N is the number of kids in the class and the kids were numbered 1 through N. snoopy and flymouse were always numbered 1 and N. Then follow M lines each holding three integers AB and c in order, meaning that kid A believed that kid Bshould never get over c candies more than he did.

    Output

    Output one line with only the largest difference desired. The difference is guaranteed to be finite.

    Sample Input

    2 2
    1 2 5
    2 1 4

    Sample Output

    5

    Hint

    32-bit signed integer type is capable of doing all arithmetic.

    Source


            先输入n,m

            接下来m行,每行输入A,B,C

            输入A B C,表示孩子B最多比孩子A多C块蛋糕,问孩子1与孩子N最多相差多少块蛋糕!

    Tips:

                    不能用queue<>队列来做,模拟队列可以!



    #include "stdio.h"
    #include "string.h"
    //#include "queue"
    //using namespace std;
    
    #define N 30005  //图中点的个数
    #define INF 0x3fffffff
    
    struct node
    {
        int x,y;
        int weight;
        int next;
    }edge[5*N];
    
    int n,m;
    int dist[N];
    bool mark[N];
    int head[N],idx;
    int stack[5*N];
    
    void Init();
    void SPFA();
    void Add(int x,int y,int k);
    
    int main()
    {
        int i;
        int x,y,k;
        scanf("%d %d",&n,&m);
            Init();
            for(i=0; i<m; ++i)
            {
                scanf("%d %d %d",&x,&y,&k); /***y-x<=k***从点x到点y建边,权值为k**/
                Add(x,y,k);
            }
            SPFA();
            printf("%d
    ",dist[n]);
        return 0;
    }
    
    void Init()
    {
        idx = 0;
        memset(head,-1,sizeof(head));
    }
    
    void Add(int x,int y,int k)
    {
        edge[idx].x = x;
        edge[idx].y = y;
        edge[idx].weight = k;
        edge[idx].next = head[x];
        head[x] = idx++;
    }
    
    void SPFA()  //这题只能模拟队列~
    {
        int i;
        int x,y;
        memset(mark,false,sizeof(mark));
        for(i=1; i<=n; ++i)  dist[i] = INF;
        int top = 0;
        //queue<int> q;
        stack[++top] = 1;//q.push(1);
        mark[1] = true;
        dist[1] = 0;
        while(top)
        {
            x = stack[top--];
            //q.pop();
            for(i=head[x]; i!=-1; i=edge[i].next)
            {
                y = edge[i].y;
                if(dist[y] > dist[x] + edge[i].weight)
                {
                    dist[y] = dist[x] + edge[i].weight;
                    if(!mark[y])
                    {
                        mark[y] = true;
                        stack[++top] = y;//q.push(y);
                    }
                }
            }
            mark[x] = false;
        }
    }


  • 相关阅读:
    操作系统精髓读书笔记
    springboot 项目中读取资源文件内容 如图片、文档文件
    重构-改善既有代码的设计读书小结
    投资中最简单的事读书笔记
    公司的行业差异
    Linux-TCP之深入浅出send和recv
    Linux-socket的close和shutdown区别及应用场景
    C-pthread_cond_wait 详解
    Linux-文件描述符的本质及与文件指针的区别
    数据结构-树的进化及与数据库的关系
  • 原文地址:https://www.cnblogs.com/ruo-yu/p/4411980.html
Copyright © 2011-2022 走看看