学步园

最基础的单源最短路径

dijkstra 复杂度O(n^2)

#include<cstring>
#include<string>
#include<fstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cctype>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<stack>
#include<ctime>
#include<cstdlib>
#include<functional>
#include<cmath>
using namespace std;
#define PI acos(-1.0)
#define MAXN 50100
#define eps 1e-7
#define INF 0x7FFFFFFF
#define long long ll;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

int edge[110][110];
int vis[110];
int dist[110];
int n,m;
void dijkstra(int v){
    int i,j;
    for(i=1;i<=n;i++){
        dist[i] = edge[v][i];
    }
    vis[v] = 1;
    for(i=0;i<n-1;i++){
        int temp = INF;
        int k = -1;
        for(j=1;j<=n;j++){
            if(!vis[j]&&dist[j]<temp){
                temp = dist[j];
                k = j;
            }
        }
        if(k==-1)   break;
        vis[k] = 1;
        for(j=1;j<=n;j++){
            if(!vis[j]&&dist[k]!=INF&&edge[k][j]!=INF&&dist[j]>dist[k]+edge[k][j])
                dist[j] = dist[k] + edge[k][j];
        }
    }
}
int main(){
    int i,j,a,b,c;
    while(scanf("%d%d",&n,&m),n||m){
        for(i=1;i<=n;i++){
            for(j=1;j<=n;j++){
                edge[i][j] = INF;
            }
        }
        memset(vis,0,sizeof(vis));
        for(i=0;i<m;i++){
            scanf("%d%d%d",&a,&b,&c);
            edge[a][b] = edge[b][a] = c;
        }
        dijkstra(1);
        printf("%d\n",dist[n]);
    }
    return 0;
}

Floyd 复杂度O(n^3)

#include<stdio.h>
#include<cstring>
#include<algorithm>
#include<math.h>
#include<iostream>
using namespace std;

int main()
{
    int n,m,a[105][105],i,j,k,x,y,t;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==0&&m==0)
            return 0;
        memset(a,0,sizeof(a));
        for(i=0;i<m;i++)
        {
            scanf("%d%d%d",&x,&y,&t);
            a[x][y]=t;
            a[y][x]=t;
        }
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
                for(k=1;k<=n;k++)
                {
                    if(a[i][j]&&a[i][k]&&(a[j][k]>a[i][j]+a[i][k]||a[j][k]==0))
                        a[j][k]=a[i][j]+a[i][k];
                }
        printf("%d\n",a[1][n]);
    }
    return 0;
}