hdu1125--附上多个最短路方法代码

最近学习最短路算法，发现这个题目基本上能用很多最短路方法解，于是贴出所有代码：spfa邻接矩阵、spfa--数组邻接表、dijikstra、floyd、bellman_ford

spfa--邻接矩阵版

// AC 176k 0ms#include<cstdio>#include<queue>#include<cstring>using namespace std;#define MAX 10//想不通的是本来这里我用10 去调试..可以提交的时候忘了改..也过了..无语int N;//N个经纪人int v[MAX][MAX];//邻接表v[i][j]表示i号经纪人人到j号经纪人的时间int d[MAX];//d[i]表示源点(也就是每次起点)到i的最少时间bool flag[MAX];//每次spfa 对于每一个点入队一次，标记入队void Input(){int m,to,time;//每个人的m个朋友，以及用时memset(v,0,sizeof(v));for(int i=1;i<=N;i++){scanf("%d",&m);while(m--){scanf("%d%d",&to,&time);v[i][to]=time;}}}void spfa(){int i,j;int ans[MAX];for(i=1;i<=N;i++) ans[i]=0;for(i=1;i<=N;i++)//1到N都spfa一次{for(j=1;j<=N;j++) d[j]=1001,flag[j]=false;//这里之所以d[i]初始化为1001，是因为100个人，每个人传达10 应该不会超过1001d[i]=0;flag[i]=true;queue<int> q;q.push(i);while(!q.empty()){int u=q.front();q.pop();for(j=1;j<=N;j++)if(v[u][j]!=0 && d[j]>d[u]+v[u][j])// && !flag[j]这里注意...不管j在不在队列，都要松弛一次{d[j]=d[u]+v[u][j];//松弛if(!flag[j]){flag[j]=true;q.push(j);}}}for(j=1;j<=N;j++)//一次spfa之后 d[]最大的就是本次的i到所有点的最短用时，即最短路ans[i]= ans[i]>d[j] ? ans[i]:d[j];}int ANS1=1001,ANS2;for(i=1;i<=N;i++)//取每个起点的最短路中 最小的值if(ANS1>ans[i])ANS1=ans[i],ANS2=i;    if(ANS1!=1001)    printf("%d %d\n",ANS2,ANS1);}int main(){while(scanf("%d",&N),N){Input();spfa();}return 0;}

spfa数组邻接表

// AC 180k 16ms#include<cstdio>#include<queue>using namespace std;#define MAX 1001struct node{int to,time;int next;}edge[MAX*2];int head[MAX];int tol;void add(int st,int end,int time){edge[tol].to=end;edge[tol].time=time;edge[tol].next=head[st];head[st]=tol++;}int N;void init(){int i;for(i=1;i<=N;i++) head[i]=-1;tol=0;for(i=1;i<=N;i++){int m,to,time;scanf("%d",&m);while(m--){scanf("%d%d",&to,&time);add(i,to,time);}}}int d[MAX];bool flag[MAX];int spfa(int s){int i;for(i=1;i<=N;i++) flag[i]=false, d[i]=10*MAX;d[s]=0;queue<int>q;q.push(s);while(!q.empty()){int u=q.front();q.pop(); flag[u]=false;for(int j=head[u];j!=-1;j=edge[j].next){int v=edge[j].to,w=edge[j].time;if(d[v]>d[u]+w){d[v]=d[u]+w;if(!flag[v]) flag[v]=true,q.push(v);}}}int ans=0;for(i=1;i<=N;i++)if(d[i]>ans) ans=d[i];return ans;}int  main(){while(scanf("%d",&N),N){init();int ans=10*MAX,ans_start;for(int i=1;i<=N;i++){int x=spfa(i);if(x<ans)ans=x,ans_start=i;}printf("%d %d\n",ans_start,ans);}return 0;}

floyd--不用说肯定是邻接矩阵：

// 168k 0ms#include<stdio.h>#define MAX 6#define INF 1<<29int N;int map[MAX][MAX];void init(){int i,to,m,time;for(i=1;i<=N;i++)for(int j=1;j<=N;j++)map[i][j]=INF;for(i=1;i<=N;i++){scanf("%d",&m);while(m--){scanf("%d%d",&to,&time);map[i][to]=time;}}}void floyd(){int i,j,k;for(k=1;k<=N;k++)for(i=1;i<=N;i++)for(j=1;j<=N;j++)if(i!=j) map[i][j]=map[i][j]>map[i][k]+map[k][j] ? map[i][k]+map[k][j]:map[i][j];    int ans1,ans2=INF;for(i=1;i<=N;i++){int max=0;for(j=1;j<=N;j++)if(i!=j && map[i][j]>max) max=map[i][j];if(max<ans2) ans2=max,ans1=i;}printf("%d %d\n",ans1,ans2);}int main(){while(scanf("%d",&N),N){init();floyd();}return 0;}

dijikstra-邻接矩阵

#include<stdio.h>#define MAX 100int N;int map[MAX][MAX];void init(){int i,j;for(i=1;i<=N;i++)for(j=1;j<=N;j++)map[i][j]=0;int m,to,time;for(i=1;i<=N;i++){scanf("%d",&m);while(m--){scanf("%d%d",&to,&time);map[i][to]=time;}}}int d[MAX];bool flag[MAX];int Dijkstra(int s){int i,j;for(i=1;i<=N;i++) flag[i]=false,d[i]=1001;d[s]=0;for(i=1;i<=N;i++){int u=0,max=1001;for(j=1;j<=N;j++){if(!flag[j] && max>d[j])u=j,max=d[j];}if(u==0) break;flag[u]=true;for(j =1; j<=N;j++)if(map[u][j]!=0 && d[j]>d[u]+map[u][j])d[j]=d[u]+map[u][j];}int ans=0;for(i=1;i<=N;i++)ans=ans>d[i] ? ans:d[i];return ans;}int main(){while(scanf("%d",&N),N){init();int ans1=1001,ans2;for(int i=1;i<=N;i++){   int x=Dijkstra(i);   if(ans1>x)   ans1=x,ans2=i;}printf("%d %d\n",ans2,ans1);}return 0;}

bellman_ford--结构体数组

//AC 168k 0ms#include<stdio.h>#define MAX 101struct node{int st,end,time;}edge[MAX];int tol;int N;void init(){int i;tol=0;int m,to,time;for(i=1;i<=N;i++){scanf("%d",&m);while(m--){scanf("%d%d",&to,&time);edge[tol].st=i,edge[tol].end=to,edge[tol++].time=time;}}}int d[MAX];int Bellman_ford(int s){int i;for(i=1;i<=N;i++) d[i]=1001;d[s]=0;for(i=1;i<N;i++){for(int j=0;j<tol;j++){int u=edge[j].st,v=edge[j].end,w=edge[j].time;d[v]=d[u]+w<d[v] ? d[u]+w:d[v];//松弛}}int ans=0;for(i=1;i<=N;i++)ans=ans>d[i] ? ans:d[i];return ans;}int main(){while(scanf("%d",&N),N){init();int ans1=1001,ans2;for(int i=1;i<=N;i++){int x=Bellman_ford(i);   if(ans1 > x)   ans1=x,ans2=i;}printf("%d %d\n",ans2,ans1);}return 0;}