（poj）3159 Candies

题目链接：http://poj.org/problem?id=3159

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   and   respectively. N is the number of kids in the class and the kids were numbered  through N. snoopy and flymouse were always numbered  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

Sample Output

题意：flymouse是幼稚园班上的班长，一天老师给小朋友们买了一堆的糖果，由flymouse来分发。flymouse希望自己分得的糖果数尽量多于snoopy。对于其他小朋友而言，则必须自己得到的糖果不少于班上某某，给出m个这种约束关系（u，v, w）即同学u的糖果数不能比同学v的糖果数少w。问flymouse最多能多snoopy几个糖果。

方法：用spfa写，开始用队列存，提交超时了，然后用堆栈存，测试数据问题

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <stack>
 #include <queue>
 #include <math.h>
 #include <vector>
 using namespace std;
 #define N  30010
 #define ll long long
 #define INF 0x3f3f3f3f
 #define met(a,b) memset(a,b,sizeof(a));
 vector<vector<int> >Q;
 struct node
 {
     int v,l,next;

 }s[];
 int Map[N],vis[N],dis[N];
 int k,n,q[N];
 void add(int e,int f,int l)
 {
     s[k].l=l;
     s[k].next=Map[e];
     s[k].v=f;
     Map[e]=k++;
 }
 void spfa()
 {
     int t=;
     met(dis,);
     for(int i=;i<=n;i++)
         dis[i]=INF;
     q[++t]=;///用堆栈存
     vis[]=;
     dis[]=;
     while(t)
     {
         int u=q[t--];
         vis[u]=;
         for(int i=Map[u];i!=-;i=s[i].next)
         {
             int v=s[i].v;
             if(dis[v]>dis[u]+s[i].l)
             {
                 dis[v]=dis[u]+s[i].l;
                 if(!vis[v])
                 {
                     q[++t]=v;
                     vis[v]=;
                 }
             }
         }
     }
 }
 int main()
 {
     int m,e,f,l;
     while(scanf("%d %d",&n,&m)!=EOF)
     {
         met(Map,-);k=;
         for(int i=;i<m;i++)
         {
             scanf("%d %d %d",&e,&f,&l);
             add(e,f,l);
         }
         spfa();
         printf("%d\n",dis[n]);
     }
     return ;
 }