[SinGuLaRiTy] the shortest-circuit computing code base, singularity code base

[SinGuLaRiTy] the shortest-circuit computing code base, singularity code base

[SinGuLaRiTy-1002]Copyright (c) SinGuLaRiTy 2017. All Rights Reserved.

Dijkstra:

Description

 

Single-Source Vertex Shortest Path of Directed Graph (Dijkstra algorithm)

 

Input

 

Row 1st: an integer n (2 <= n <= 500) and m (10 <= m <= 20000) separated by two spaces ), the number of vertices and edges of the graph.

2nd. m + 1 rows: three separate spaces in each row are integers I, j, and w. I indicates the start point of an edge, j indicates the end point, and w indicates the weight value.

Row m + 2: 2 integers s, t (1 <= s, t <= n), indicating the specified vertex.

 

Output

 

Row 3: Minimum Distance

Row 2nd: Shortest Path (the sequence from the start point to the end point, separated by one space)

 

 

Sample Input	Sample output
5 7 1 2 10 1 4 30 1 5 100 2 3 50 3 5 10 4 3 20 4 5 60 1 5	60 1 4 3 5

 

 

 

 

 

 

 

 

 

Code:

#include<iostream>#include<cstring>#include<cstdlib>#include<cstdio>#include<cmath>using namespace std;int n,m;int G[501][501],dis[501],L[501];bool used[501];void Dijkstra(int s){    for(int i=1;i<=n;i++)    {        used[s]=1;        for(int j=1;j<=n;j++)        {            if(G[s][j]==0||s==j)                continue;            if(dis[j]>dis[s]+G[s][j])                dis[j]=dis[s]+G[s][j],L[j]=s;        }        int themin=0x3f3f3f3f;        for(int i=1;i<=n;i++)                            if(used[i]==0&&dis[i]<themin)                s=i,themin=dis[i];    }}int main(){    int a,b,c,S,E,cnt=1;    scanf("%d%d",&n,&m);    for(int i=1;i<=m;i++)    {        scanf("%d%d%d",&a,&b,&c);        G[a][b]=c;    }    scanf("%d%d",&S,&E);    int e=E;    dis[S]=0;    for(int i=1;i<=n;i++)        if(i!=S) dis[i]=0x3f3f3f3f;                    Dijkstra(S);          printf("%d\n",dis[E]);    while(L[E]!=0)    {        dis[cnt++]=L[E];        E=L[E];    }    for(int i=cnt-1;i>=1;i--)        printf("%d ",dis[i]);    printf("%d",e);    return 0;}

 

SPAF:

Description

The shortest path of a single-source vertex in a directed graph. The Source Vertex number is 1, and the end vertex number is n.

Input

Row 1st: an integer n (2 <= n <= 5000) and m (10 <= m <= 500000) separated by two spaces ), the number of vertices and edges of the graph.

2nd. m + 1 rows: three separate spaces in each row are integers I, j, and w. I indicates the start point of an edge, j indicates the end point, and w indicates the weight value.

Output

Row 1st: 1 integer, indicating the minimum distance

Example input 1	Sample output 1
4 71 2 681 3 191 4 662 3 233 4 653 2 574 1 68	66

 

 

 

 

 

Sample Input	Sample output
3 31 2-72 3 43 1 2	No Solution

 

 

 

 

Code:

#include<iostream>#include<cstdio>#include<cstring>#include<queue>using namespace std;struct node{    int v,w,next;}edge[500050];int cnt,head[5050],n,m,dis[5050],tot[5050];bool inq[5050];void addedge(int u,int v,int w){    edge[cnt].v=v;    edge[cnt].w=w;    edge[cnt].next=head[u];    head[u]=cnt++;}bool SPFA(){    memset(dis,0x3f,sizeof dis);    dis[1]=0;    inq[1]=1;    deque<int>q;    q.push_front(1);    while(!q.empty())    {        int u=q.front();        q.pop_front();        inq[u]=0;        for(int i=head[u];i!=-1;i=edge[i].next)        {            int v=edge[i].v,w=edge[i].w;            if(dis[v]>dis[u]+w)            {                dis[v]=dis[u]+w;                if(!inq[v])                {                    if(dis[v]<dis[u])q.push_front(v);                    else q.push_back(v);                    inq[v]=1;                    if(++tot[v]>n)return 0;                }            }        }    }    return 1;}int main(){    int u,v,w;    scanf("%d%d",&n,&m);    memset(head,-1,sizeof head);    for(int i=1;i<=m;++i)    {        scanf("%d%d%d",&u,&v,&w);        addedge(u,v,w);    }    SPFA()?printf("%d\n",dis[n]):    puts("No Solution");}

 

Floyd:

Code:

#include<cstdio>  #include<cstring>  #define maxn 100  #define INF -1  int map[maxn][maxn];  int n, m, path[maxn][maxn];    void Floyd(int n)  {      int i, j, k;      for(k = 0; k < n; ++k)          for(i = 0; i < n; ++i)              for(j = 0; j < n; ++j)                  if(map[i][k] != INF && map[k][j] != INF && (map[i][k] + map[k][j] < map[i][j] || map[i][j] == INF)){                      map[i][j] = map[i][k] + map[k][j];                      path[i][j] = k;                  }             }    void getPath(int v, int u)  {      int k = path[v][u];      if(k == INF){          printf("%d===>", v);                  return;      }      getPath(v, k);      getPath(k, u);  }    int main()  {       scanf("%d%d", &n, &m);      memset(map, INF, sizeof(map));      memset(path, INF, sizeof(path));      int i, j, a, b, c;      for(i = 0; i < n; ++i) map[i][i] = 0;      for(i = 0; i < m; ++i){          scanf("%d%d%d", &a, &b, &c);          map[a][b] = c;      }      Floyd(n);      for(i = 0; i < n; ++i)          for(j = 0; j < n; ++j)              if(map[i][j] != INF){                  printf("%d->%d:%d\n the path is:", i, j, map[i][j]);                  getPath(i, j);                  printf("%d\n", j);              }      return 0;  }  

 

BellmanFord:

Description

The single-Source Vertex shortest path with negative weight (BellmanFord algorithm) of a directed graph. If there are multiple shortest paths, the output path goes through a smaller number of edges. If the shortest path has the same number of edges, output a sequence with a smaller number.

Input

Row 1st: an integer n (2 <= n <= 500) and m (10 <= m <= 20000) separated by two spaces ), the number of vertices and edges of the graph.

2nd. m + 1 rows: three separate spaces in each row are integers I, j, and w. I indicates the start point of an edge, j indicates the end point, and w indicates the weight value.

Row m + 2: 2 integers s, t (1 <= s, t <= n), indicating the specified vertex.

Output

Row 3: Minimum Distance

Row 2nd: Shortest Path (the sequence from the start point to the end point, separated by one space)

If a negative weight loop appears, the output is No Solution.

Sample Input	Sample output
6 7 1 2 2 1 3-1 2 4-3 3 4 3 3 6 7 4 6-2 3 5 6 1 6	-3 1 2 4 6

 

 

 

 

 

 

Code:

#include<cstdio> #include<cstring> #include<algorithm> using namespace std; int f[20005][3],dis[505],s[505]; void print(int x){    if(s[x]==0) return;    print(s[x]);    printf(" %d",x);}int main() {    int n,m,i,j,q,z;     scanf("%d%d",&n,&m);     for(i=1;i<=m;i++)        scanf("%d%d%d",&f[i][0],&f[i][1],&f[i][2]);     scanf("%d%d",&q,&z);    memset(dis,1,sizeof(dis));     dis[q]=0;    for(i=1;i<=n;i++){        for(j=1;j<=m;j++){            if(dis[f[j][0]]+f[j][2]<dis[f[j][1]]&&i<n){                dis[f[j][1]]=dis[f[j][0]]+f[j][2];                s[f[j][1]]=f[j][0];            }            if(i==n&&dis[f[j][0]]+f[j][2]<dis[f[j][1]]){                printf("No Solution");                return 0;            }        }    }    printf("%d\n",dis[z]);     printf("%d",q);     print(z);}

 

Time: