POJ 3159 Candies 【差分约束】

最新推荐文章于 2022-04-27 15:11:27 发布

原创最新推荐文章于 2022-04-27 15:11:27 发布 · 207 阅读

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本文解决了一个有趣的问题：如何在满足所有孩子的满意度条件下，使两个特定孩子之间的糖果数量差距最大化。通过使用SPFA算法并结合栈来优化计算过程。

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Candies

Time Limit: 1500MS		Memory Limit: 131072K
Total Submissions: 36441		Accepted: 10243

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 Mlines 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.

spfa中要用stack，用queue会超时，不知为什么，如果哪位大佬知道，可否留言讲解一波

#include<queue>
#include<cstdio>
#include<cstring>
#include<stack>
#include<iostream>
using namespace std;
const int MAX = 1000000;
const int INF = 0x3f3f3f3f;
struct node{
    int to, c, next;
    node() {}
    node(int to, int c, int next) : to(to), c(c), next(next) {}
} e[MAX];
int cnt = 0, n, m;
int head[MAX], dis[MAX], vis[MAX];
stack <int> q;
void add(int x, int y, int z){
    e[cnt] = {y, z, head[x]};
    head[x] = cnt++;
}
void spfa(){
    q.push(1);
    memset(vis, 0, sizeof vis);
    memset(dis, INF, sizeof dis);
    vis[1] = 1;
    dis[1] = 0;
    while(!q.empty()){
        int u = q.top();
        vis[u] = 0;
        q.pop();
        for(int i = head[u]; i != -1; i = e[i].next){
            int to = e[i].to;
            int c = e[i].c;
            if(dis[to] > dis[u] + c){
                dis[to] = dis[u] + c;
                if(!vis[to]){
                    q.push(to);
                    vis[to] = 1;
                }
            }
        }
    }
}
int main(){
    scanf("%d%d", &n, &m);
    memset(head, -1, sizeof head);
    for(int i = 0, x, y, z; i < m; i++){
        scanf("%d%d%d", &x, &y, &z);
        add(x, y, z);
    }
    spfa();
    printf("%d\n", dis[n]);
    return 0;
}