洛谷P3371 【模板】单源最短路径（弱化版） #最短路 SPFA算法 / Dijkstra算法链式前向星#

本文链接： https://blog.csdn.net/qq_35850147/article/details/88533218

题目背景

本题测试数据为随机数据，在考试中可能会出现构造数据让SPFA不通过，如有需要请移步 P4779。

题目描述

如题，给出一个有向图，请输出从某一点出发到所有点的最短路径长度。

输入输出格式

输入格式：

第一行包含三个整数N、M、S，分别表示点的个数、有向边的个数、出发点的编号。

接下来M行每行包含三个整数Fi、Gi、Wi，分别表示第i条有向边的出发点、目标点和长度。

输出格式：

一行，包含N个用空格分隔的整数，其中第i个整数表示从点S出发到点i的最短路径长度（若S=i则最短路径长度为0，若从点S无法到达点i，则最短路径长度为2147483647）

输入输出样例

输入样例#1： 复制

输出样例#1： 复制

0 2 4 3

说明

时空限制：1000ms,128M

数据规模：

对于20%的数据：N<=5，M<=15；

对于40%的数据：N<=100，M<=10000；

对于70%的数据：N<=1000，M<=100000；

对于100%的数据：N<=10000，M<=500000。保证数据随机。

对于真正 100% 的数据，请移步 P4779。请注意，该题与本题数据范围略有不同。

样例说明：

图片1到3和1到4的文字位置调换

题解

Version 1 SPFA

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e4 + 10;
const int maxm = 5e5 + 10;
const int inf = 2147483647;
bool vis[maxn];
int n, m, s, cnt, head[maxn], dis[maxn];
struct edge { int v, w, next; } e[maxm];

inline const int read()
{
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = (x << 3) + (x << 1) + ch - '0'; ch = getchar(); }
    return x * f;
}

void addedge(int u, int v, int w)
{
    e[cnt].v = v;
    e[cnt].w = w;
    e[cnt].next = head[u];
    head[u] = cnt++;
}

void spfa(int src)
{
    for (int i = 1; i <= n; i++) dis[i] = inf;
    dis[s] = 0;
    queue<int> q;
    q.push(s);
    while (!q.empty())
    {
        int u = q.front(); q.pop();
        vis[u] = false;
        for (int i = head[u]; ~i; i = e[i].next)
        {
            int v = e[i].v, w = e[i].w;
            if (dis[v] > dis[u] + w)
            {
                dis[v] = dis[u] + w;
                if (!vis[v])
                {
                    q.push(v);
                    vis[v] = true;
                }
            }
        }
    }
}

int main()
{
    n = read(); m = read(); s = read();
    memset(head, -1, sizeof(head));
    while (m--)
    {
        int u = read(), v = read(), w = read();
        addedge(u, v, w);
    }
    spfa(s);
    for (int i = 1; i <= n; i++) printf("%d%s", dis[i], i == n ? "\n" : " ");
    return 0;
}

Version 2 Dijkstra朴素版本

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e4 + 10;
const int maxm = 5e5 + 10;
const int inf = 2147483647;
bool vis[maxn];
int n, m, s, cnt, head[maxn], dis[maxn];
struct edge { int v, w, next; } e[maxm];

inline const int read()
{
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = (x << 3) + (x << 1) + ch - '0'; ch = getchar(); }
    return x * f;
}

void addedge(int u, int v, int w)
{
    e[cnt].v = v;
    e[cnt].w = w;
    e[cnt].next = head[u];
    head[u] = cnt++;
}

void dijkstra(int src)
{
    for (int i = 1; i <= n; i++) dis[i] = inf;
    dis[src] = 0;
    for (int j = 0; j < n - 1; j++)
    {
        int u = 0, d = inf;
        for (int i = 1; i <= n; i++)
        {
            if (!vis[i] && dis[i] < d)
            {
                d = dis[i];
                u = i;
            }
        }
        vis[u] = true;
        for (int i = head[u]; ~i; i = e[i].next)
        {
            int v = e[i].v, w = e[i].w;
            dis[v] = min(dis[v], dis[u] + w);
        }
    }
}

int main()
{
    n = read(); m = read(); s = read();
    memset(head, -1, sizeof(head));
    while (m--)
    {
        int u = read(), v = read(), w = read();
        addedge(u, v, w);
    }
    dijkstra(s);
    for (int i = 1; i <= n; i++) printf("%d%s", dis[i], i == n ? "\n" : " ");
    return 0;
}

Version 3 Dijkstra堆优化版本

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e4 + 10;
const int maxm = 5e5 + 10;
const int inf = 2147483647;
bool vis[maxn];
int n, m, s, cnt, head[maxn], dis[maxn];
struct edge { int v, w, next; } e[maxm];
struct node
{
    int u, w;
    bool operator < (const node &n) const { return w > n.w; }
};
priority_queue<node> q;

inline const int read()
{
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = (x << 3) + (x << 1) + ch - '0'; ch = getchar(); }
    return x * f;
}

void addedge(int u, int v, int w)
{
    e[cnt].v = v;
    e[cnt].w = w;
    e[cnt].next = head[u];
    head[u] = cnt++;
}

void dijkstra(int src)
{
    for (int i = 1; i <= n; i++) dis[i] = inf;
    dis[src] = 0;
    q.push(node{src, 0});
    while (!q.empty())
    {
        int u = q.top().u; q.pop();
        if (vis[u]) continue;
        vis[u] = true;
        for (int i = head[u]; ~i; i = e[i].next)
        {
            int v = e[i].v, w = e[i].w;
            if (dis[v] > dis[u] + w)
            {
                dis[v] = dis[u] + w;
                q.push(node{v, dis[v]});
            }
        }
    }
}

int main()
{
    n = read(); m = read(); s = read();
    memset(head, -1, sizeof(head));
    while (m--)
    {
        int u = read(), v = read(), w = read();
        addedge(u, v, w);
    }
    dijkstra(s);
    for (int i = 1; i <= n; i++) printf("%d%s", dis[i], i == n ? "\n" : " ");
    return 0;
}