解題關鍵：最短路的變形。

1、按頂點存儲，$O(n^2)$

#include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<iostream> #include<cmath> #include<vector> #define inf 0x3f3f3f3f #define max_v 502 using namespace std; typedef long long ll; int pre[max_v],pathnum[max_v],toval[max_v],val[max_v]; int cost[max_v][max_v],d[max_v],used[max_v],V; void dijkstra(int s){ //邊權和點權都是通過頂點體現(xiàn)fill(d,d+V,inf);fill(used,used+V,0);fill(pre,pre+V,-1);d[s]=0;pathnum[s]=1;toval[s]=val[s];while(true){int v=-1;for(int u=0;u<V;u++){if(!used[u]&&(v==-1||d[u]<d[v])) v=u;}if(v==-1) break;used[v]=true;for(int u=0;u<V;u++){if(used[u]==false&&cost[u][v]!=inf){//????if(d[u]>d[v]+cost[v][u]){d[u]=d[v]+cost[v][u];pre[u]=v;pathnum[u]=pathnum[v];toval[u]=toval[v]+val[u];}else if(d[u]==d[v]+cost[v][u]){pathnum[u]+=pathnum[v];if(toval[u]<toval[v]+val[u]){pre[u]=v;toval[u]=toval[v]+val[u];}}}}} } vector<int> get_path(int t){vector<int>path;for(;t!=-1;t=pre[t]) path.push_back(t);reverse(path.begin(), path.end());return path; }int N,M,S,D; int main(){cin>>N>>M>>S>>D;V=N;for(int i=0;i<N;i++) cin>>val[i];for(int i=0;i<V;i++){for(int j=0;j<V;j++){if(i!=j) cost[i][j]=cost[i][j]=inf;}}for(int i=0;i<M;i++){int a,b,c;cin>>a>>b>>c;cost[a][b]=cost[b][a]=c;}dijkstra(S);printf("%d %d\n",pathnum[D],toval[D]);vector<int>path=get_path(D);for(int i=0;i<path.size();i++){printf("%d%c",path[i],i==path.size()-1?'\n':' ');}return 0; }

?2、按邊存儲 $O(nlogn)$

#include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<cmath> #include<iostream> #include<vector> #include<queue> #define max_v 502 #define inf 0x3f3f3f3f using namespace std; typedef long long ll; struct edge{int to,cost; }; typedef pair<int,int>P;//按邊存儲，默認不可以自己到自己進行dp int V,pre[max_v],pathnum[max_v],val[max_v],toval[max_v]; vector<edge>G[max_v]; int d[max_v]; void dijkstra(int s){priority_queue<P,vector<P>,greater<P> >que;fill(d,d+V,inf);fill(pre,pre+V,-1);toval[s]=val[s];pathnum[s]=1;d[s]=0;que.push(P(0,s));while(!que.empty()){P p=que.top();que.pop();int v=p.second;if(d[v]<p.first) continue;for(int i=0;i<G[v].size();i++){edge e=G[v][i];if(d[e.to]>d[v]+e.cost){d[e.to]=d[v]+e.cost;pathnum[e.to]=pathnum[v];toval[e.to]=toval[v]+val[e.to];pre[e.to]=v;que.push(P(d[e.to],e.to));}else if(d[e.to]==d[v]+e.cost){pathnum[e.to]+=pathnum[v];if(toval[e.to]<toval[v]+val[e.to]){toval[e.to]=toval[v]+val[e.to];pre[e.to]=v;}}}} }vector<int> get_path(int t){vector<int>path;for(;t!=-1;t=pre[t]) path.push_back(t);reverse(path.begin(), path.end());return path; }int N,M,S,D; int main(){cin>>N>>M>>S>>D;V=N;for(int i=0;i<N;i++) cin>>val[i];for(int i=0;i<M;i++){int a,b,c;cin>>a>>b>>c;G[a].push_back((edge){b,c});G[b].push_back((edge){a,c});}dijkstra(S);printf("%d %d\n",pathnum[D],toval[D]);vector<int>path=get_path(D);for(int i=0;i<path.size();i++){printf("%d%c",path[i],i==path.size()-1?'\n':' ');}return 0; }

?3、代碼三

#include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<iostream> #include<cmath> #include<vector> #define inf 0x3f3f3f3f #define max_v 502 using namespace std; typedef long long ll; int pre[max_v],pathnum[max_v],toval[max_v],val[max_v]; int cost[max_v][max_v],d[max_v],used[max_v],V; void dijkstra(int s){ //邊權和點權都是通過頂點體現(xiàn)fill(d,d+V,inf);fill(used,used+V,0);fill(pre,pre+V,-1);d[s]=0;pathnum[s]=1;toval[s]=val[s];while(true){int v=-1;for(int u=0;u<V;u++){if(!used[u]&&(v==-1||d[u]<d[v])) v=u;}if(v==-1) break;used[v]=true;for(int u=0;u<V;u++){if(d[u]>d[v]+cost[v][u]){d[u]=d[v]+cost[v][u];pre[u]=v;pathnum[u]=pathnum[v];toval[u]=toval[v]+val[u];}else if(d[u]==d[v]+cost[v][u]){pathnum[u]+=pathnum[v];if(toval[u]<toval[v]+val[u]){pre[u]=v;toval[u]=toval[v]+val[u];}}}} } vector<int> get_path(int t){vector<int>path;for(;t!=-1;t=pre[t]) path.push_back(t);reverse(path.begin(), path.end());return path; }int N,M,S,D; int main(){cin>>N>>M>>S>>D;V=N;for(int i=0;i<N;i++) cin>>val[i];for(int i=0;i<V;i++){for(int j=0;j<V;j++){cost[i][j]=cost[i][j]=inf;}}for(int i=0;i<M;i++){int a,b,c;cin>>a>>b>>c;cost[a][b]=cost[b][a]=c;}dijkstra(S);printf("%d %d\n",pathnum[D],toval[D]);vector<int>path=get_path(D);for(int i=0;i<path.size();i++){printf("%d%c",path[i],i==path.size()-1?'\n':' ');}return 0; }

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轉載于:https://www.cnblogs.com/elpsycongroo/p/8512203.html

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