最短路 P1144 最短路计数Dijkstra堆优化/SPFA

Posted 2021-02-19 jason66661010

题目

https://www.luogu.com.cn/problem/P1144

题目分析

注意相同距离的最短路径条数的判断！（使用数组！）

代码

Dijkstra+堆优化

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#define maxn 1000002
#define maxm 2000002
#define inf 0x3f3f3f3f
using namespace std;
struct edge
{
    int to;
    int next;
    int dis;

}e[maxm];
struct node
{
    int dis;

    int pos;
};
bool operator<(const node &a, const node &b)

{
    return a.dis>b.dis;

}
int head[maxn], dist[maxn], cnt, visited[maxn],counts[maxn];

int n, m;
void addedge(int u, int v, int w)
{
    cnt++;
    e[cnt].to = v;
    e[cnt].dis = w;
    e[cnt].next = head[u];
    head[u] = cnt;
}
priority_queue<node>q;
void Dijkstra(int s)
{
    fill(dist, dist + n + 1, inf);
    dist[s] = 0; counts[s] = 1;
    q.push({0,s});
    while (!q.empty())
    {
        node tmp = q.top(); q.pop();
        int x = tmp.pos;
        if (visited[x])continue;
        visited[x] = 1;
        for (int i = head[x]; i; i = e[i].next)
        {
            int y = e[i].to;
            if (dist[y] > dist[x] + e[i].dis)

            {
                dist[y] = dist[x] + e[i].dis;

                counts[y] = counts[x];

                q.push({ dist[y], y });
            }
            else if (dist[y] == dist[x] + e[i].dis)

            {
                counts[y] += counts[x];

                counts[y] %= 100003;
            }
        }
    
    }
}
int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 0; i < m; i++)
    {
        int a, b;

        scanf("%d%d", &a, &b);

        addedge(a, b, 1);
        addedge(b, a, 1);
    }
    Dijkstra(1);

    for (int i = 1; i <= n; i++)
        printf("%d
", counts[i]);


}

SPFA

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#define maxn 1000002
#define maxm 2000002
#define inf 0x3f3f3f3f
using namespace std;
struct edge
{
    int to;
    int next;
    int dis;
}e[maxm];
int head[maxn], dist[maxn], cnt, visited[maxn], counts[maxn];
int n, m;
void addedge(int u, int v, int w)
{
    cnt++;
    e[cnt].to = v;
    e[cnt].dis = w;
    e[cnt].next = head[u];
    head[u] = cnt;
}
queue<int>q;
int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 0; i < m; i++)
    {
        int a, b;
        scanf("%d%d", &a, &b);
        addedge(a, b, 1);
        addedge(b, a, 1);
    }
    fill(dist, dist + n + 1, inf);
    dist[1] = 0; counts[1] = 1;
    visited[1] = 1;
    q.push(1);
    while (!q.empty())
    {
        int x = q.front(); q.pop();
        visited[x] = 0;//这里将visited[x]置为0，防止下面干扰能使路径变短（dist[y] > dist[x] + 1）且不在队列里的节点的判断
        for (int i = head[x]; i; i = e[i].next)
        {
            int y = e[i].to;
            if (dist[y] > dist[x] + 1)//能使路径变短（dist[y] > dist[x] + 1）且不在队列里的节点才入队
            {
                dist[y] = dist[x] + 1;
                counts[y] = counts[x];
                    if (!visited[y])//节点重复入队是没有意义的！！
                {
                    q.push(y);
                    visited[y] = 1;
                }
            }
            else if (dist[y] == dist[x] + 1)
            {
                counts[y] += counts[x];
                counts[y] %= 100003;
            }
        }
    }
    for (int i = 1; i <= n; i++)
        printf("%d
", counts[i]);
}