链接:
http://acm.hdu.edu.cn/showproblem.php?pid=2544
题目:
Problem Description
在每年的校赛里,所有进入决赛的同学都会获得一件很漂亮的t-shirt。但是每当我们的工作人员把上百件的衣服从商店运回到赛场的时候,却是非常累的!所以现在他们想要寻找最短的从商店到赛场的路线,你可以帮助他们吗?
  
Input
输入包括多组数据。每组数据第一行是两个整数N、M(N<=100,M<=10000),N表示成都的大街上有几个路口,标号为1的路口是商店所在地,标号为N的路口是赛场所在地,M则表示在成都有几条路。N=M=0表示输入结束。接下来M行,每行包括3个整数A,B,C(1<=A,B<=N,1<=C<=1000),表示在路口A与路口B之间有一条路,我们的工作人员需要C分钟的时间走过这条路。
 输入保证至少存在1条商店到赛场的路线。
Output
对于每组输入,输出一行,表示工作人员从商店走到赛场的最短时间
Sample Input
   2 1 1 2 3 3 3 1 2 5 2 3 5 3 1 2 0 0
Sample Output
   3 2
Source
UESTC 6th Programming Contest Online
基础最短路,不解释,其实是专门用来验证各种最短路模板的。
1. Dijkstra 普通版
   -  #include<cstdio>
-  #include<cstring>
-  const int N=105, INF=9999999;
-  int d[N], w[N][N],vis[N],n,m;
-  void Dijkstra(int src){
-  for(int i=1; i<=n; ++i)
-  d[i] = INF;
-  d[src] = 0;
-  memset(vis, 0, sizeof(vis));
-  for(int i=1; i<=n; ++i){
-  int u=-1;
-  for(int j=1; j<=n; ++j)if(!vis[j]){
-  if(u==-1 || d[j]<d[u]) u=j;
-  }
-  vis[u] = 1;
-  for(int j=1; j<=n; ++j)if(!vis[j]){
-  int tmp = d[u] + w[u][j];
-  if(tmp<d[j]) d[j] = tmp;
-  }
-  }
-  }
-  int main(){
-  int a,b,c;
-  while(~scanf("%d%d",&n,&m)&&n+m){
-  for(int i=1; i<=n; ++i){
-  w[i][i] = INF;
-  for(int j=i+1; j<=n; ++j)
-  w[i][j] = w[j][i] = INF;
-  }
-  for(int i=0; i<m; ++i){
-  scanf("%d%d%d",&a,&b,&c);
-  w[a][b] = w[b][a] = c;
-  }
-  Dijkstra(1);
-  printf("%d\n", d[n]);
-  }
-  return 0;
-  }
2. Dijkstra+邻接表(用数组实现)+优先队列优化
   -  #include<cstdio>
-  #include<cstring>
-  #include<utility>
-  #include<queue>
-  using namespace std;
-  const int N=20005;
-  const int INF=9999999;
-  typedef pair<int,int>pii;
-  priority_queue<pii, vector<pii>, greater<pii> >q;
-  int d[N], first[N], u[N], v[N], w[N], next[N],n,m;
-  bool vis[N];
-  // 无向图的输入,注意每输入的一条边要看作是两条边
-  void read_graph(){
-  memset(first, -1, sizeof(first)); //初始化表头
-  for(int e=1; e<=m; ++e){
-  scanf("%d%d%d",&u[e], &v[e], &w[e]);
-  u[e+m] = v[e]; v[e+m] = u[e]; w[e+m] = w[e]; // 增加一条它的反向边
-  next[e] = first[u[e]]; // 插入链表
-  first[u[e]] = e;
-  next[e+m] =first[u[e+m]]; // 反向边插入链表
-  first[u[e+m]] = e+m;
-  }
-  }
-  void Dijkstra(int src){
-  memset(vis, 0, sizeof(vis));
-  for(int i=1; i<=n; ++i) d[i] = INF;
-  d[src] = 0;
-  q.push(make_pair(d[src], src));
-  while(!q.empty()){
-  pii u = q.top(); q.pop();
-  int x = u.second;
-  if(vis[x]) continue;
-  vis[x] = true;
-  for(int e = first[x]; e!=-1; e=next[e]) if(d[v[e]] > d[x]+w[e]){
-  d[v[e]] = d[x] + w[e];
-  q.push(make_pair(d[v[e]], v[e]));
-  }
-  }
-  }
-  int main(){
-  int a,b,c;
-  while(~scanf("%d%d",&n,&m)&&n+m){
-  read_graph();
-  Dijkstra(1);
-  printf("%d\n", d[n]);
-  }
-  return 0;
-  }
3. Dijkstra+邻接表(用vecor实现)+优先队列优化
   -  #include<cstdio>
-  #include<cstring>
-  #include<utility>
-  #include<queue>
-  #include<vector>
-  using namespace std;
-  const int N=105;
-  const int INF=9999999;
-  typedef pair<int,int>pii;
-  vector<pii>G[N];
-  priority_queue<pii, vector<pii>, greater<pii> >q;
-  int d[N], first[N], u[N], v[N], w[N], next[N],n,m;
-  bool vis[N];
-  // 无向图的输入,注意没输入的一条边要看作是两条边
-  void read_graph(){
-  for(int i=1; i<=n; ++i)
-  G[i].clear();
-  int a,b,c;
-  for(int i=1; i<=m; ++i){
-  scanf("%d%d%d",&a,&b,&c);
-  G[a].push_back(make_pair(b,c));
-  G[b].push_back(make_pair(a,c));
-  }
-  }
-  void Dijkstra(int src){
-  memset(vis, 0, sizeof(vis));
-  for(int i=1; i<=n; ++i) d[i] = INF;
-  d[src] = 0;
-  q.push(make_pair(d[src], src));
-  while(!q.empty()){
-  pii t = q.top(); q.pop();
-  int u = t.second;
-  if(vis[u]) continue;
-  vis[u] = true;
-  for(int v=0; v<G[u].size(); ++v)if(d[G[u][v].first] > d[u]+G[u][v].second){
-  d[G[u][v].first] = d[u]+G[u][v].second;
-  q.push(make_pair(d[G[u][v].first], G[u][v].first));
-  }
-  }
-  }
-  int main(){
-  int a,b,c;
-  while(~scanf("%d%d",&n,&m)&&n+m){
-  read_graph();
-  Dijkstra(1);
-  printf("%d\n", d[n]);
-  }
-  return 0;
-  }
  
二,Bellman-Ford算法
   -  #include<cstdio>
-  #include<cstring>
-  #include<utility>
-  #include<queue>
-  using namespace std;
-  const int N=20005;
-  const int INF=9999999;
-  int n, m, u[N],v[N],w[N], d[N];
-  // 无向图的输入,注意每输入的一条边要看作是两条边
-  inline void read_graph(){
-  for(int e=1; e<=m; ++e){
-  scanf("%d%d%d",&u[e],&v[e],&w[e]);
-  }
-  }
-  inline void Bellman_Ford(int src){
-  for(int i=1; i<=n; ++i) d[i] = INF;
-  d[src] = 0;
-  for(int k=0; k<n-1; ++k){
-  for(int i=1; i<=m; ++i){
-  int x=u[i], y=v[i];
-  if(d[x] < INF){
-  if(d[y]>d[x]+w[i])
-  d[y] = d[x]+w[i];
-  }
-  if(d[y] < INF){
-  if(d[x]>d[y]+w[i])
-  d[x] = d[y]+w[i];
-  }
-  }
-  }
-  }
-  int main(){
-  int a,b,c;
-  while(~scanf("%d%d",&n,&m)&&n+m){
-  read_graph();
-  Bellman_Ford(1);
-  printf("%d\n", d[n]);
-  }
-  return 0;
-  }
三,SPFA
邻接表实现
   -  #include<cstdio>
-  #include<cstring>
-  #include<utility>
-  #include<queue>
-  using namespace std;
-  const int N=20005;
-  const int INF=2147483646>>1;
-  int n, m, first[N],next[N],u[N],v[N],w[N], d[N];
-  bool vis[N];
-  queue<int>q;
-  inline void read_graph(){
-  memset(first, -1, sizeof(first));
-  for(int e=1; e<=m; ++e){
-  scanf("%d%d%d",&u[e],&v[e],&w[e]);
-  u[e+m]=v[e], v[e+m]=u[e], w[e+m]=w[e];
-  next[e] = first[u[e]];
-  first[u[e]] = e;
-  next[e+m] = first[u[e+m]];
-  first[u[e+m]] = e+m;
-  }
-  }
-  void SPFA(int src){
-  memset(vis, 0, sizeof(vis));
-  for(int i=1; i<=n; ++i) d[i] = INF;
-  d[src] = 0;
-  vis[src] = true;
-  q.push(src);
-  while(!q.empty()){
-  int x = q.front(); q.pop();
-  vis[x] = false;
-  for(int e=first[x]; e!=-1; e=next[e]){
-  if(d[x]+w[e] < d[v[e]]){
-  d[v[e]] = d[x]+w[e];
-  if(!vis[v[e]]){
-  vis[v[e]] = true;
-  q.push(v[e]);
-  }
-  }
-  }
-  }
-  }
-  int main(){
-  int a,b,c;
-  while(~scanf("%d%d",&n,&m)&&n+m){
-  read_graph();
-  SPFA(1);
-  printf("%d\n", d[n]);
-  }
-  return 0;
-  }
四, Floyd算法
   -  #include<cstdio>
-  #include<cstring>
-  #include<utility>
-  #include<queue>
-  using namespace std;
-  const int N=105;
-  const int INF=2147483646;
-  int n, m, d[N][N];
-  inline void read_graph(){
-  for(int i=1; i<=n; ++i){
-  d[i][i] = INF;
-  for(int j=i+1; j<=n; ++j)
-  d[i][j]=d[j][i]=INF;
-  }
-  int a,b,c;
-  for(int e=1; e<=m; ++e){
-  scanf("%d%d%d",&a,&b,&c);
-  d[a][b]=d[b][a]=c;
-  }
-  }
-  inline void Floyd(int src){
-  for(int k=1; k<=n; ++k){
-  for(int i=1; i<=n; ++i){
-  for(int j=1; j<=n; ++j)
-  if(d[i][k]<INF && d[k][j]<INF){ //防止溢出
-  d[i][j] = min(d[i][j], d[i][k]+d[k][j]);
-  }
-  }
-  }
-  }
-  int main(){
-  int a,b,c;
-  while(~scanf("%d%d",&n,&m)&&n+m){
-  read_graph();
-  Floyd(1);
-  printf("%d\n", d[1][n]);
-  }
-  return 0;
-  }
   —— 生命的意义,在于赋予它意义。
原创 http://blog.csdn.net/shuangde800 , By D_Double (转载请标明)

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京公网安备 11010502036488号