最短路劲

HDU-2544-最短路

http://acm.hdu.edu.cn/showproblem.php?pid=2544

 赤裸裸的Dijkstra

Dijkstra算法是典型的单源最短路径算法，用于计算一个节点到其他所有节点的最短路径

#include<stdio.h>#include<string.h>#include<stdlib.h>#define maxvalue 0x7fffffffint map[105][105];int n,m;int dijkstra(){int i,j,v,temp;int dis[105];int visit[105];for(i=1;i<=n;i++)dis[i]=map[1][i];dis[1]=0;memset(visit,0,sizeof(visit));visit[1]=1;for(i=1;i<n;i++){temp=maxvalue;v=1;for(j=1;j<=n;j++)if(visit[j]==0&&dis[j]<temp){v=j;temp=dis[j];}visit[v]=1;for(j=1;j<=n;j++)if(visit[j]==0&&map[v][j]<maxvalue&&dis[j]>dis[v]+map[v][j])  //少了map[v][j]<maxvalue会WAdis[j]=dis[v]+map[v][j];}return dis[n];}int main(){int i,j,ans;int a,b,c;while(scanf("%d %d",&n,&m),n||m){for(i=1;i<=n;i++)for(j=1;j<=n;j++)map[i][j]=maxvalue;while(m--){scanf("%d %d %d",&a,&b,&c);if(map[a][b]>c)   //防止有重边{map[a][b]=map[b][a]=c;}}ans=dijkstra();printf("%d\n",ans);}return 0;}

这题用floyd也能做

Floyd是通过一个图的邻接矩阵求出它的每两点间的最短路径矩阵，时间复杂度为O(n^3)

#include<stdio.h>#include<string.h>#include<stdlib.h>#define maxvalue 0x7fffffffint map[105][105];int n,m;int floyd(){int i,j,k;for(k=1;k<=n;k++)for(i=1;i<=n;i++)for(j=1;j<=n;j++){if(map[i][k]<maxvalue&&map[k][j]<maxvalue&&map[i][k]+map[k][j]<map[i][j])map[i][j]=map[i][k]+map[k][j];}return map[1][n];}int main(){int i,j,ans;int a,b,c;while(scanf("%d %d",&n,&m),n||m){for(i=1;i<=n;i++)for(j=1;j<=n;j++)map[i][j]=maxvalue;while(m--){scanf("%d %d %d",&a,&b,&c);if(map[a][b]>c)   //防止有重边{map[a][b]=map[b][a]=c;}}ans=floyd();printf("%d\n",ans);}return 0;}

用dijkstra和floyd也可以求出最短路劲的序列，即从始点至终点要依次经过哪些点

比如说求如下图，从V1至V10的最短路劲序列

输入邻接矩阵，-1表示两点之间没有路劲

dijkstra

#include<stdio.h>#include<string.h>#include<stdlib.h>#define maxint 0x7fffffffint map[15][15];int dis[15];int visit[15];int prev[15];void print(int x){    int k;k=prev[x];if(k==1)return;print(k);printf(" %d",k);}void dijkstra(){int i,j,v,min;memset(visit,0,sizeof(visit));visit[1]=1;for(i=1;i<=10;i++){        dis[i]=map[1][i];if(dis[i]==maxint)prev[i]=0;elseprev[i]=1;}for(i=1;i<10;i++){min=maxint;for(j=1;j<=10;j++)        {if(!visit[j]&&dis[j]<min){min=dis[j];v=j;}}visit[v]=1;for(j=1;j<=10;j++)if(!visit[j]&&map[v][j]!=maxint&&dis[j]>map[v][j]+dis[v])        {dis[j]=map[v][j]+dis[v];prev[j]=v;}}}int main(){int i,j;for(i=1;i<=10;i++)for(j=1;j<=10;j++){scanf("%d",&map[i][j]);if(map[i][j]==-1)map[i][j]=maxint;}dijkstra();printf("1");print(10);printf(" 10\n");    return 0;}

floyd

#include<stdio.h>#include<string.h>#include<stdlib.h>int dis[15][15];int path[15][15];void floyd(){int i,j,k;memset(path,0,sizeof(path));for(k=1;k<=10;k++)for(i=1;i<=10;i++)for(j=1;j<=10;j++)if(dis[i][k]!=-1&&dis[k][j]!=-1&&(dis[i][j]==-1||dis[i][j]>dis[i][k]+dis[k][j])){dis[i][j]=dis[i][k]+dis[k][j];path[i][j]=k;}}void print(int i,int j){    int k;k=path[i][j];if(k==0)return;print(i,k);printf(" %d",k);print(k,j);}int main(){int i,j;for(i=1;i<=10;i++)for(j=1;j<=10;j++)scanf("%d",&dis[i][j]);floyd();printf("1");print(1,10);printf(" 10\n");return 0;}

结果均为1 3 5 8 10

POJ-3268-Silver Cow Party

http://poj.org/problem?id=3268

很有意思的一题，题意是： n头牛要去参加一场在编号为x的牛的农场举行的派对,每头牛都必须参加完派对后回到家，每头牛都会选择最短路径，求这n个牛的最短路径（一个来回）中最长的一条的长度。

dijkstra用于计算一个节点到其他所有节点的最短路径，将邻接矩阵转置，可求得其他所有节点到该节点的最短路劲

先求出x点到各点的最短路径放到数组dis1中，再将邻接矩阵转置，求出各点到x点的最短路径放到数组dis2中，求出max(dis1 + dis2)即可

#include<stdio.h>#include<string.h>#include<stdlib.h>#define max 0x7fffffint map[1005][1005];int dis1[1005];int dis2[1005];int n,m,x;void dijkstra(int dis[]){int i,j,v,min;int visit[1005];memset(visit,0,sizeof(visit));for(i=1;i<=n;i++)dis[i]=map[x][i];dis[x]=0;visit[x]=1;for(i=1;i<=n;i++){min=max;for(j=1;j<=n;j++){if(!visit[j]&&dis[j]<min){min=dis[j];v=j;}}visit[v]=1;for(j=1;j<=n;j++)    if(!visit[j]&&dis[v]+map[v][j]<dis[j])dis[j]=dis[v]+map[v][j];}}void tran(){int i,j,temp;for(i=1;i<=n;i++)for(j=1;j<=i;j++){temp=map[i][j];map[i][j]=map[j][i];map[j][i]=temp;}}int main(){int a,b,c,i,j;int ans;while(scanf("%d%d%d",&n,&m,&x)!=EOF){for(i=1;i<=n;i++)for(j=1;j<=n;j++)map[i][j]=max;while(m--){scanf("%d%d%d",&a,&b,&c);if(map[a][b]>c)map[a][b]=c;}dijkstra(dis1);  //x点到各点的最短路径tran();dijkstra(dis2);  //各点到x点的最短路径ans=-1;for(i=1;i<=n;i++){if(i!=x&&dis1[i]+dis2[i]>ans)ans=dis1[i]+dis2[i];}printf("%d\n",ans);}return 0;}