SPFA template vector, spfavector

SPFA template vector, spfavector

Shortest Path for algorithm training
Time Limit: 1.0 s memory limit: 256.0 MB

Problem description

Given n vertices and Directed Graphs of m edges (some edge weights may be negative, but there is no negative ring ). Calculate the shortest path (from vertex 1 to vertex n) from vertex 1 to other ).

Input Format

The first line has two integers, n and m.
In the next m row, each row has three integers u, v, l, indicating that u to v has an edge with a length of l.

Output Format

There are n-1 rows in total. line I indicates the shortest path from Point 1 to point I + 1.

Sample Input

3 3
1 2-1
2 3-1
3 1 2

Sample output

-1
-2

Data scale and conventions

For 10% of the data, n = 2, m = 2.

For 30% of the data, n <= 5, m <= 10.

For 100% of data, 1 <= n <= 20000,1 <= m <= 200000,-10000 <= l <= 10000, to ensure that all other vertices can be reached from any vertex.

Question link: http://lx.lanqiao.org/problem.page? Gpid = T15

Question Analysis: Repeat the vector board of SPFA.

#include <cstdio>#include <cstring>#include <queue>#include <vector>using namespace std;int const MAX = 200005;int const INF = 1 << 30;int n, m;struct EDGE{    int u, v;    int val;}e[MAX << 2];struct NODE{    int v, w;    NODE(int vv, int ww)    {        v = vv;        w = ww;    }};vector <NODE> vt[MAX];int dis[MAX];bool vis[MAX];void SPFA(int v0){    memset(vis, false, sizeof(vis));    for(int i = 1; i <= n; i++)        dis[i] = INF;    dis[v0] = 0;    queue <int> q;    q.push(v0);    while(!q.empty())    {        int u = q.front();        q.pop();        vis[u] = false;        int sz = vt[u].size();        for(int i = 0; i < sz; i++)        {            int v = vt[u][i].v;            int w = vt[u][i].w;            if(dis[v] > dis[u] + w)            {                dis[v] = dis[u] + w;                if(!vis[v])                {                    q.push(v);                    vis[v] = true;                }            }        }    }}int main(){    scanf("%d %d", &n, &m);    for(int i = 0; i < m; i++)        scanf("%d %d %d", &e[i].u, &e[i].v, &e[i].val);    for(int i = 0; i < m; i++)        vt[e[i].u].push_back(NODE(e[i].v, e[i].val));    SPFA(1);    for(int i = 2; i <= n; i++)        printf("%d\n", dis[i]);}