POJ 3159 Candies（差分约束）

原创于 2018-11-19 20:06:46 发布 · 162 阅读

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#差分约束

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本文详细解析了POJ3159 Candies问题，这是一个经典的差分约束系统问题，涉及到小朋友们分糖时的嫉妒心限制。通过构建图模型和使用SPFA算法求解最长路径，最终找到在满足所有条件下的最大糖果分配差距。

POJ 3159 Candies

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

题意：有n个小朋友，你现在要给这些小朋友们分发糖果，但小朋友们之间有嫉妒心。接下来m行，每行三个数，分别表示小朋友A希望小朋友B得到的糖果不能比他多x个。要求你计算在满足所有小朋友的条件的情况下最多需要准备多少颗糖。

题解：num[B]<=num[A]+x，很明显是个差分约束问题。求最大值的话就是跑1-n的最短路，对应求最小值就是跑1-n的最长路。

此题用队列的话会T掉，可以改成栈或者想办法优化。

#include<iostream>
#include<cstdio>
#include<stack>
#include<cstring>
using namespace std;
const int maxn=300000+10;
struct cc{
    int from,to,cost;
}es[maxn];
int first[maxn];
int nxt[maxn];
bool vis[maxn];
int dis[maxn];
stack<int>q;
int tot=0;
void build(int ff,int tt,int pp)
{
    es[++tot]=(cc){ff,tt,pp};
    nxt[tot]=first[ff];
    first[ff]=tot;
}
void spfa(int s)
{
    q.push(s);
    vis[s]=1;
    dis[s]=0;
    while(!q.empty())
    {
        int u=q.top(); q.pop();
        vis[u]=0;
        for(int i=first[u];i;i=nxt[i])
        {
            int v=es[i].to;
            if(dis[v]>dis[u]+es[i].cost)
            {
                dis[v]=dis[u]+es[i].cost;
                if(!vis[v])
                {
                    q.push(v);
                    vis[v]=1;
                }
            }
        }
    }
}
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        int x,y,z;
        scanf("%d%d%d",&x,&y,&z);
        build(x,y,z);
    }
    memset(dis,63,sizeof(dis));
    spfa(1);
    printf("%d",dis[n]);
    return 0;
}