HDOJ 1874 畅通工程续(Dijkstra+Floyed+SPFA)

畅通工程续

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 23601    Accepted Submission(s): 8310

Problem Description

某省自从实行了很多年的畅通工程计划后，终于修建了很多路。不过路多了也不好，每次要从一个城镇到另一个城镇时，都有许多种道路方案可以选择，而某些方案要比另一些方案行走的距离要短很多。这让行人很困扰。

现在，已知起点和终点，请你计算出要从起点到终点，最短需要行走多少距离。

 

Input

本题目包含多组数据，请处理到文件结束。
每组数据第一行包含两个正整数N和M(0<N<200,0<M<1000)，分别代表现有城镇的数目和已修建的道路的数目。城镇分别以0～N-1编号。
接下来是M行道路信息。每一行有三个整数A,B,X(0<=A,B<N,A!=B,0<X<10000),表示城镇A和城镇B之间有一条长度为X的双向道路。
再接下一行有两个整数S,T(0<=S,T<N)，分别代表起点和终点。

 

Output

对于每组数据，请在一行里输出最短需要行走的距离。如果不存在从S到T的路线，就输出-1.

 

Sample Input

3 30 1 10 2 31 2 10 23 10 1 11 2

 

Sample Output

2-1

 

Author

linle

Dijkstra(优先队列)

#include<iostream>#include<cstring>#include<cstdlib>#include<functional>#include<vector>#include<queue>#define MAX 1005#define INF 0x3f3f3f3fusing namespace std;struct Edge{//每条边的属性 int from, to, cost;Edge(int f, int t, int c) :from(f), to(t), cost(c){}};int N, M;vector<Edge>edge[MAX];//创建存储边的数组 int Mincost[MAX];typedef pair<int, int >P;//将最短的边的长度，跟点通过pair作为一个整体   注意：长度做first，节点做second void  Dijkstra(int s){priority_queue<P, vector<P>, greater<P> >pq;//最小值优先出对的优先队列 fill(Mincost, Mincost + MAX, INF);//将所有的点的最短距离初始化为INF //fill(Mincost,Mincost+M,INF);Mincost[s] = 0; pq.push(P(0, s));//起点的最短距离设为0，将(0,s)压入 while (!pq.empty()){P temp = pq.top(); pq.pop();int tempnode = temp.second;//tempnode  接受点的信息 if (Mincost[tempnode] != temp.first) continue;//如果这个点的最短距离不等于.first  继续循环 for (int i = 0; i<edge[tempnode].size(); i++){//依次遍历以这个点为起点的边 Edge e = edge[tempnode][i];if (Mincost[e.to]>Mincost[tempnode] + e.cost){//如果某点的距离因为别的点变小，则改变(比如，某点开始时无法到达起点，但后来因为某点加入导致距离变小，则需改变) Mincost[e.to] = Mincost[tempnode] + e.cost;pq.push(P(Mincost[e.to], e.to));//压入改变的点 }}}}int main(){while (cin >> N >> M){memset(edge, 0, sizeof(edge));for (int i = 0; i<M; i++){int A, B, C; cin >> A >> B >> C;Edge e(A, B, C);Edge e2(B, A, C);//无向图，各为起点 edge[A].push_back(e);//以A为起点的边 edge[B].push_back(e2);//以B为起点的边 }int s, t; cin >> s >> t;Dijkstra(s);if (Mincost[t] != INF)cout << Mincost[t] << endl;else cout << -1 << endl;}}

Djikstra

#include<iostream>#include<cstdlib>#include<cstring>#define INF 0x3f3f3f3f#define N 205using namespace std;int n, m;int Map[N][N], Mincost[N];int visit[N];void Init(){for (int i = 0; i < n; i++){for (int j = 0; j < n; j++){Map[i][j] = INF;}}}void Dijkstra(int s){int Min, post;memset(visit, 0, sizeof(visit));//所有点都为选过 for (int i = 0; i < n; i++){Mincost[i] =  Map[s][i];//与s相接的点的长度 }visit[s] = 1; Mincost[s] = 0;//s标记，及本点的距离为0 for (int i = 0; i<n; i++){Min = INF;for (int j = 0; j<n; j++){if (!visit[j] && Min>Mincost[j]){Min = Mincost[j];post = j;}}if (Min != INF){visit[post] = 1;for (int j = 0; j<n; j++){if (!visit[j] && Mincost[j]>Mincost[post] + Map[post][j]){//更改距离 Mincost[j] = Mincost[post] + Map[post][j];}}}}}int main(){while (cin >> n >> m){Init();for (int i = 0; i<m; i++){int from, to, cost;cin >> from >> to >> cost;if (Map[from][to]>cost){//很关键，在此wa了好几次，，作用：防止重边Map[from][to] = cost;Map[to][from] = cost;}}int s, t; cin >> s >> t;Dijkstra(s);if (s == t) cout << 0 << endl;else if (Mincost[t] != INF){cout << Mincost[t] << endl;}else cout << -1 << endl;}}

Floyd

#include<iostream>#include<cstdlib>#include<cstring>#define MAX 205#define INF 10000005using namespace std;int n,m,Map[MAX][MAX];void Init(){for(int i=0;i<n;i++){for(int j=0;j<n;j++){Map[i][j]=(i==j)?0:INF;}} }void Floyd(){for(int k=0;k<n;k++){for(int i=0;i<n;i++){for(int j=0;j<n;j++){//if(Map[i][j]<INF && Map[k][j]<INF)Map[i][j]=min(Map[i][j],Map[i][k]+Map[k][j]);}}}}int main(){while(cin>>n>>m){Init();for(int i=0;i<m;i++){int a,b,x;cin>>a>>b>>x;if(Map[a][b]>x){Map[a][b]=Map[b][a]=x;}}int s,t; cin>>s>>t;Floyd();if(Map[s][t]<INF){cout<<Map[s][t]<<endl;}else cout<<-1<<endl;}} 

SPFA

#include<iostream>#include<cstring>#include<cstdlib>#include<queue>#define MAX 205#define INF 0x3f3f3f3fusing namespace std; int Map[MAX][MAX],Mincost[MAX],visit[MAX];int n,m;void Inite(){for(int i=0;i<n;i++){for(int j=0;j<n;j++){Map[i][j]=INF;}}}void SPFA(int s){for(int i=0;i<n;i++){Mincost[i]=INF;}memset(visit,0,sizeof(visit));Mincost[s]=0,visit[s]=1;queue<int>q; q.push(s);while(!q.empty()){int temp=q.front(); q.pop();visit[temp]=0;for(int i=0;i<n;i++){if(Mincost[i]>Mincost[temp]+Map[temp][i]){Mincost[i]=Mincost[temp]+Map[temp][i];if(!visit[i]){visit[i]=1;q.push(i); }}}}}int main(){while(cin>>n>>m){Inite();for(int i=0;i<m;i++){int a,b,c; cin>>a>>b>>c;if(Map[a][b]>c){Map[a][b]=Map[b][a]=c;}}int s,t; cin>>s>>t;SPFA(s);if(Mincost[t]!=INF){cout<<Mincost[t]<<endl;}else cout<<-1<<endl;}}

SPFA (邻接表)

#include<iostream>#include<cstdlib>#include<cstring>#include<vector>#include<queue>#define MAX 205#define MAX_W 1005#define INF 0x3f3f3f3fusing namespace std;struct Edge{int from, to, cost;Edge(int f, int t, int c) :from(f), to(t), cost(c){}};vector<Edge>edge[MAX_W];int Mincost[MAX];int n, m, visit[MAX];void SPFA(int s){for (int i = 0; i<MAX; i++){Mincost[i] = INF;}memset(visit, 0, sizeof(visit));visit[s] = 1, Mincost[s] = 0;queue<int>q; q.push(s);while (!q.empty()){int tempnode = q.front(); q.pop();visit[tempnode] = 0;for (int i = 0; i<edge[tempnode].size(); i++){Edge tempE = edge[tempnode][i];int id = tempE.to;if (Mincost[id]>Mincost[tempnode] + tempE.cost){Mincost[id] = Mincost[tempnode] + tempE.cost;if (!visit[id]){visit[id] = 1;q.push(id);}}}}}int main(){while (cin >> n >> m){memset(edge, 0, sizeof(edge));for (int i = 0; i<m; i++){int a, b, c; cin >> a >> b >> c;Edge e(a, b, c);Edge e2(b, a, c);edge[a].push_back(e);edge[b].push_back(e2);}int s, t; cin >> s >> t;SPFA(s);if (Mincost[t] != INF){cout << Mincost[t] << endl;}else cout << -1 << endl;}}