【SPFA】poj1511

Invitation Cards

Description

In the age of television, not many people attend theater performances. Antique Comedians of Malidinesia are aware of this fact. They want to propagate theater and, most of all, Antique Comedies. They have printed invitation cards with all the necessary information and with the programme. A lot of students were hired to distribute these invitations among the people. Each student volunteer has assigned exactly one bus stop and he or she stays there the whole day and gives invitation to people travelling by bus. A special course was taken where students learned how to influence people and what is the difference between influencing and robbery. 

The transport system is very special: all lines are unidirectional and connect exactly two stops. Buses leave the originating stop with passangers each half an hour. After reaching the destination stop they return empty to the originating stop, where they wait until the next full half an hour, e.g. X:00 or X:30, where 'X' denotes the hour. The fee for transport between two stops is given by special tables and is payable on the spot. The lines are planned in such a way, that each round trip (i.e. a journey starting and finishing at the same stop) passes through a Central Checkpoint Stop (CCS) where each passenger has to pass a thorough check including body scan. 

All the ACM student members leave the CCS each morning. Each volunteer is to move to one predetermined stop to invite passengers. There are as many volunteers as stops. At the end of the day, all students travel back to CCS. You are to write a computer program that helps ACM to minimize the amount of money to pay every day for the transport of their employees. 

Input

The input consists of N cases. The first line of the input contains only positive integer N. Then follow the cases. Each case begins with a line containing exactly two integers P and Q, 1 <= P,Q <= 1000000. P is the number of stops including CCS and Q the number of bus lines. Then there are Q lines, each describing one bus line. Each of the lines contains exactly three numbers - the originating stop, the destination stop and the price. The CCS is designated by number 1. Prices are positive integers the sum of which is smaller than 1000000000. You can also assume it is always possible to get from any stop to any other stop.

Output

For each case, print one line containing the minimum amount of money to be paid each day by ACM for the travel costs of its volunteers.

Sample Input

2
2 2
1 2 13
2 1 33
4 6
1 2 10
2 1 60
1 3 20
3 4 10
2 4 5
4 1 50

Sample Output

46
210

大意:求源点到每个点的距离和加上每个点到原点的距离和
思路,先正向一次SPFA再讲图反转进行一次SPFA
WA了三次,然后就看了一下discuss发现要用longlong才行……

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;;
const long N = 1000010, M = 1000010;
const long INF = 1073741824;
struct ty
{
    long t, w, next;
};
long tt, n, t, m;
long long sum;
long head1[N], head2[N];
ty edge1[M], edge2[M];
long q[1000000];
bool v[N];
long dist[N];
void insertedge1(long x, long y, long z, long k)
{
    edge1[k].t = y;
    edge1[k].w = z;
    edge1[k].next = head1[x];
    head1[x] = k;
}
void insertedge2(long x, long y, long z, long k)
{
    edge2[k].t = y;
    edge2[k].w = z;
    edge2[k].next = head2[x];
    head2[x] = k;
}
void init()
{
    memset(head1, 0, sizeof(head1));
    memset(edge1, 0, sizeof(edge1));
    memset(head2, 0, sizeof(head2));
    memset(edge2, 0, sizeof(edge2));
    m = 0;
    sum = 0;
    for (long i = 1; i <= t; i++)
    {
        long x, y, z;
        scanf("%d%d%d", &x, &y, &z);
        insertedge1(x, y, z, ++m);
        insertedge2(y, x, z, m);
    }
}
void spfa1()
{
    memset(v, 0, sizeof(v));
    memset(dist, 127, sizeof(dist));
    long l = 0, r = 0;
    q[r++] = 1;
    v[1] = true;
    dist[1] = 0;
    while (l < r)
    {
        long x = q[l++];
        v[x] = false;
        for (long i = head1[x]; i != 0; i = edge1[i].next)
        {
            long t = edge1[i].t;
            if (dist[t] > dist[x] + edge1[i].w)
            {
                dist[t] = dist[x] + edge1[i].w;
                if (!v[t])
                {
                    q[r++] = t;
                    v[t] = true;
                }
            }
        }
    }

    for (long i = 2; i <= n; i++)
    {
        sum += dist[i];
      
    }

}

void spfa2()
{
    memset(v, 0, sizeof(v));
    memset(dist, 127, sizeof(dist));
    long l = 0, r = 0;
    q[r++] = 1;
    v[1] = true;
    dist[1] = 0;
    while (l < r)
    {
        long x = q[l++];
        v[x] = false;
        for (long i = head2[x]; i != 0; i = edge2[i].next)
        {
            long t = edge2[i].t;
            if (dist[t] > dist[x] + edge2[i].w)
            {
                dist[t] = dist[x] + edge2[i].w;
                if (!v[t])
                {
                    q[r++] = t;
                    v[t] = true;
                }
            }
        }
    }

    for (long i = 2; i <= n; i++)
        sum += dist[i];

}
int main()
{
    freopen("poj1511.in", "r", stdin);
    scanf("%d", &tt);
    for (long i = 1; i <= tt; i++)
    {
        scanf("%d%d", &n, &t);
        init();
        spfa1();
        spfa2();
        cout << sum << endl;
    }
    return 0 ;
}


### 关于 POJ 2092 的问题分析 POJ 平台上的题目通常涉及算法设计与实现,而编号为 2092 的具体题目并未在提供的引用中明确提及。然而,可以通过已知的相关资源和经验推测其可能的解决方法。 #### 差分约束系统的应用 如果假设 POJ 2092 类似于其他差分约束类问题,则可以参考类似的解决方案[^3]。这类问题的核心在于通过构建不等式组来表示变量之间的关系,并将其转化为图论中的短路径或长路径问题。例如: - 对于条件 \( P \),\( A - B = X \) 可以被分解为两个不等式: \( A - B \geq X \) 和 \( A - B \leq X \)[^3]。 - 对于条件 \( V \),则有 \( A - B \geq 1 \)。 这些不等式可以用边的形式表示在一个加权图中,随后运行 SPFA 或 Bellman-Ford 算法检测是否存在满足所有约束的解集。特别需要注意的是引入一个超级源点连接到所有节点,从而保证整个图连通性。 #### 动态规划 vs 常规方法对比 针对某些特定类型的优化问题,动态规划 (Dynamic Programming, DP) 方法能够显著提高效率并减少冗余计算量。相比之下,传统方式可能会因为重复子问题而导致性能瓶颈。尽管当前讨论未直接指向 POJ 2092 是否适用此技术路线,但从更广泛意义上看,DP 是处理复杂状态转移的有效工具之一[^2]。 以下是基于上述理论框架的一个简单 Python 实现例子用于验证可行性: ```python from collections import deque def spfa(n, edges): INF = float('inf') dist = [-INF] * n in_queue = [False] * n q = deque() s = 0 # Super source node index. # Initialize distances from super-source to all nodes as zero. for i in range(n): if not in_queue[i]: q.append(i) in_queue[i] = True while q: u = q.popleft() in_queue[u] = False for v, w in edges.get(u, []): if dist[v] < dist[u] + w: dist[v] = dist[u] + w if not in_queue[v]: q.append(v) in_queue[v] = True return any(dist[i] >= INF / 2 for i in range(1,n)) # Example usage with dummy data representing constraints... if __name__ == "__main__": N = 5 # Number of variables plus one extra 'super' start point. E = {0:[(i,0)for i in range(1,N)]} # Connects every variable directly via weight-zero links. result = spfa(N,E) print("Positive cycle exists:",result) ``` 以上代码片段展示了如何运用队列辅助广度优先搜索(SPFA)寻找潜在正向循环逻辑链路结构。这一步骤对于判定是否有可行方案至关重要。 ---
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