POJ 3159 Candies 差分约束-SPFA栈实现

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本文介绍了一个关于分配糖果的问题，通过建立差分约束系统并运用SPFA算法求解最大差值。该问题涉及多名儿童之间的糖果数量比较限制，并寻求在满足所有条件的情况下，特定两人间糖果数量的最大差异。

Candies

Time Limit: 1500MS		Memory Limit: 131072K
Total Submissions: 33931		Accepted: 9528

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

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题意：

给n个人派糖果，给出m组数据，每组数据包含A，B，C  三个数，
意思是A的糖果数比B少的个数不多于C,即B - A<= c ；
最后求n 比 1 最多多多少糖果；

学习了一下差分约束，顺便学了一下栈版本的SPFA

①：对于差分不等式，a - b <= c ，建一条 b 到 a 的权值为 c 的边，求的是最短路，得到的是最大值
②：对于不等式 a - b >= c ，建一条 b 到 a 的权值为 c 的边，求的是最长路，得到的是最小值
③：存在负环的话是无解
④：求不出最短路（dist[ ]没有得到更新）的话是任意解

AC Code：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <queue>
#include <vector>
#include <stack>
#include <iostream>
#include <algorithm>

using namespace std;

const int MAXN = 31000;
const int MAXE = 151000;
const int INF = 0x3f3f3f3f;

struct node {
    int v,w,next;
} edge[MAXE] ;

int Head[MAXN];
int Dis[MAXN];
bool vis[MAXN];
int n,m,top;

void AddEdge(int u,int v,int w) {
    edge[top].v = v;
    edge[top].w = w;
    edge[top].next = Head[u];
    Head[u] = top++;
}

void SPFA(int s) {
    stack<int> Q;
    memset(Dis,INF,sizeof(Dis));
    memset(vis,false,sizeof(vis));
    Dis[s] = 0 ; vis[s]=true;
    Q.push(s);
    while(!Q.empty()) {
        int u = Q.top(); Q.pop();
        for(int i = Head[u]; i!=-1; 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]) {
                    vis[v]=true;
                    Q.push(v);
                }
            }
        }
        vis[u] = false;
    }
}

int main() {
    int u,v,w;
    while(~scanf("%d %d",&n,&m)) {
        memset(Head,-1,sizeof(Head));
        top = 0;
        for(int i=1; i<=m; i++) {
            scanf("%d %d %d",&u,&v,&w);
            AddEdge(u,v,w);
        }
        SPFA(1);
        printf("%d\n",Dis[n]);
    }
    return 0;
}