POJ 3159-Candies（差分約束系統-SPFA+鄰接表）

kewlgrl發表於2017-04-26
Candies
Time Limit: 1500MS   Memory Limit: 131072K
Total Submissions: 31137   Accepted: 8669

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

5

Hint

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

Source

POJ Monthly--2006.12.31, Sempr

題目意思：

N個小孩子分糖果，A說B的糖果最多比自己多C個，計算N個小朋友最多相差多少糖果。

解題思路：

SPFA+鄰接表
把STL的queue改成手寫的一個陣列；
把多餘的初始化全部刪掉；
計算src=1時的dis[n]……
1500MS險過……(｡˘•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];//佇列中的節點為頂點序號
int n;//頂點個數
ArcNode * List[MAXN];//每個頂點的邊連結串列表頭指標
int inq[MAXN];//每個頂點是否在佇列中的標誌
int dist[MAXN];
int cnt[MAXN];
void SPFA(int src)
{
    int top=0;
    memset(inq,0,sizeof(inq));
    int i,u;//u為佇列頭頂點序號
    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;//構造鄰接表
        temp->weight=w;
        temp->next=NULL;
        if(List[u]==NULL) List[u]=temp;
        else
        {
            temp->next=List[u];
            List[u]=temp;
        }
    }
    SPFA(1);//源點1
    printf("%d\n",dist[n]);
    return 0;
}


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