POJ 3159-Candies（差分约束系统-SPFA+邻接表）

逃离我推掉我的手 2022-06-17 12:50 190阅读 0赞

Candies

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1500MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;131072K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;31137</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;8669</td> 
  </tr> 
 </tbody> 
</table>

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 *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

## 题目意思： ##

N个小孩子分糖果，A说B的糖果最多比自己多C个，计算N个小朋友最多相差多少糖果。

## 解题思路： ##

SPFA+邻接表  
把STL的queue改成手写的一个数组；  
把多余的初始化全部删掉；  
计算src=1时的dis\[n\]……  
1500MS险过……(｡˘•ε•˘｡)

#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <vector>
    #include <queue>
    #include <iomanip>
    #include <algorithm>
    #define MAXN 200010
    #define INF 0xfffffff
    using namespace std;
    
    struct ArcNode
    {
        int to;
        int weight;
        ArcNode *next;
    };
    int Q[MAXN];//队列中的节点为顶点序号
    int n;//顶点个数
    ArcNode * List[MAXN];//每个顶点的边链表表头指针
    int inq[MAXN];//每个顶点是否在队列中的标志
    int dist[MAXN];
    int cnt[MAXN];
    void SPFA(int src)
    {
        int top=0;
        memset(inq,0,sizeof(inq));
        int i,u;//u为队列头顶点序号
        ArcNode * temp;
        for(i=1; i<=n; ++i)//初始化
            dist[i]=INF;
        dist[src]=0;
        ++inq[src];
        Q[top++]=src;
        while(top>=0)
        {
            u=Q[--top];
            --inq[u];
            temp=List[u];
            while(temp!=NULL)
            {
                int v=temp->to;
                if(dist[v]>dist[u]+temp->weight)
                {
                    dist[v]=dist[u]+temp->weight;
                    if(!inq[v])
                    {
                        Q[top++]=v;
                        ++inq[v];
                    }
                }
                temp=temp->next;
            }
        }
    }
    int main()
    {
    #ifdef ONLINE_JUDGE
    #else
        freopen("G:/cbx/read.txt","r",stdin);
        //freopen("G:/cbx/out.txt","w",stdout);
    #endif
        int m;
        scanf("%d%d",&n,&m);
        memset(List,0,sizeof(List));
        ArcNode *temp;
        while(m--)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            temp=new ArcNode;
            temp->to=v;//构造邻接表
            temp->weight=w;
            temp->next=NULL;
            if(List[u]==NULL) List[u]=temp;
            else
            {
                temp->next=List[u];
                List[u]=temp;
            }
        }
        SPFA(1);//源点1
        printf("%d\n",dist[n]);
        return 0;
    }

[POJ Monthly--2006.12.31]: http://poj.org/searchproblem?field=source&key=POJ+Monthly--2006.12.31