【详解】Candies POJ - 3159 ⭐⭐⭐⭐【线性差分约束 链式前向星 栈优化SPFA】

Candies POJ - 3159

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.

Examples

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.

题意:

给N个小朋友分糖, 给出M组约束a, b, c表示b的糖果不能比a多c个以上, 求1号和N号的最大糖果差异数

题解:

非常显然的线性查分约束问题
对于a, b, c表示b的糖果不能比a多c个以上 , 即cnt[a]+c >= cnt[b], 可以理解为a指向b的一条权值为c的单向边.
这样一来整个系统也就转换成了图, 而对于求1和N的最大差异数, 即cnt[N]-cnt[1] 的最大值, 即1到N的最短路

求最短路时可用堆优化Dijkstra和栈优化SPFA两种

经验小结:

搞清楚差分约束的是最短路还是最长路

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <stdlib.h>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
using namespace std;
#define ms(x, n) memset(x,n,sizeof(x));
typedef  long long LL;
const int inf = 1<<30;
const LL maxn = 300010;

int N, M, edgeCnt = 0, head[maxn];
struct node{
    int to, w, next;
    node(int tt, int ww, int nn){to = tt, w = ww, next = nn;}
    node(){}
}es[maxn];
void addEdge(int u, int v, int w){
    es[edgeCnt] = node(v, w, head[u]);
    head[u] = edgeCnt++;
}

int d[maxn];
bool used[maxn];
void SPFA(int s){
    stack<int> st;      //本题卡队列
    fill(d, d+maxn, inf);
    d[1] = 0, used[1] = true;
    st.push(1);
    while(!st.empty()){
        int u = st.top();
        st.pop();
        used[u] = false;
        for(int i = head[u]; i != -1; i = es[i].next){
            int v = es[i].to, w = es[i].w;
            if(d[u]+w < d[v]){
                d[v] = d[u]+w;
                if(!used[v]){
                    st.push(v);
                    used[v] = true;
                }
            }
        }
    }
}

int main()
{
    ms(head, -1);
    int a, b, c;
    scanf("%d%d",&N,&M);
    while(M--){
        scanf("%d%d%d",&a,&b,&c);
        addEdge(a, b, c);    //差分约束转最短路
    }
    SPFA(1);
    printf("%d\n",d[N]);
	return 0;
}