poj 3159(spfa最短路径)

最新推荐文章于 2024-05-13 15:01:06 发布

my_acm

最新推荐文章于 2024-05-13 15:01:06 发布

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分类专栏：最短路径

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本文探讨了幼儿园时期糖果分配的公平性问题，飞鼠在糖果分配过程中如何通过策略使得糖果数量差异最大化，同时保持孩子们的满意度。通过输入包含学生数量、糖果比较限制的数据，求解飞鼠能够实现的最大差异。使用SPFA算法解决最短路径问题，实现高效计算。

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Candies

Time Limit: 1500MS		Memory Limit: 131072K
Total Submissions: 23152		Accepted: 6234

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 throughN. snoopy and flymouse were always numbered 1 and N. Then follow M lines each holding three integers A, B and c in order, meaning that kid A believed that kid B should 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

Hint

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

Source

POJ Monthly--2006.12.31, Sempr

题目就是求从1到n的最短路径。

AC代码：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<vector>
#include<stack>
using namespace std;
int n,m;
struct Node{
    int u,v,w;
    int next;
}Edge[150010];
int head[30010];
int k;
int spfa(){
    stack <int> st;
    int vis[30010],dis[30010];
    for(int i=1;i<=n;i++){
        dis[i]=999999;
        vis[i]=0;
    }
    dis[1]=0;
    st.push(1);
    while(!st.empty()){
        int u=st.top(); st.pop();
        vis[u]=0;

        for(int i=head[u];i;i=Edge[i].next){
            int v=Edge[i].v;
            if(dis[v]>dis[u]+Edge[i].w){
                dis[v]=dis[u]+Edge[i].w;
                if(!vis[v]){
                    st.push(v);
                    vis[v]=1;
                }
            }
        }

    }
    return dis[n];
}
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        head[i]=0;
    for(int i=1;i<=m;i++){
        int x,y,s;
        scanf("%d%d%d",&x,&y,&s);
        k++;
        Edge[k].u=x; Edge[k].v=y; Edge[k].w=s;
        Edge[k].next=head[x];
        head[x]=k;
    }
    printf("%d\n",spfa());
    return 0;
}