HDU-3499Flight (分层图dijkstra)
一开始想的并查集(我一定是脑子坏掉了),晚上听学姐讲题才知道就是dijkstra两层;
题意:有一次机会能使一条边的权值变为原来的一半,询问从s到e的最短路。
将dis数组开成二维,第一维表示从源点到点i的路径长度,第二维表示是否使用了该次机会,并以此不断更新。
用map将字符串转为int,剩下的就是dijkstra+优先队列;
#include<cstdio> #include<cstring> #include<iostream> #include<map> #include<queue> #define ll long long #define INF 1e18 using namespace std; const int N = 1e5+10; const int M = 5e5+10; int tot,m,n; struct Edge { int v, cost, next; }edge[M]; struct qnode { int loc,level; ll dis; bool operator < (const qnode &t) const { return dis > t. dis; } qnode(int a=0,int b=0,ll c=0){ loc=a,level=b,dis=c; } }d[N][2]; int head[M]; bool vis[N][2]; void addedge(int u, int v, ll w) { edge[tot]. v = v; edge[tot]. cost = w; edge[tot]. next = head[u]; head[u] = tot ++; } void dijkstra(int st) { memset(vis, 0, sizeof(vis)); for(int i=1;i<=n;i++){ for(int j=0;j<=1;j++){ d[i][j].dis=INF; } } d[st][0].dis = 0; priority_queue <qnode> q; q. push(qnode(st,0,0)); while(! q. empty()){ qnode t = q. top(); q. pop(); int t1=t.loc,t2=t.level; if(vis[t1][t2])continue; vis[t1][t2]=true; for(int i = head[t1]; i != -1; i = edge[i]. next){ int v = edge[i]. v; int cost = edge[i]. cost; if( d[v][t2].dis > d[t1][t2].dis + cost){ d[v][t2].dis = d[t1][t2].dis + cost; q. push(qnode(v,t2,d[v][t2].dis)); } if(t2==0&&d[v][t2+1].dis>d[t1][t2].dis+cost/2){ //分别记录每一段是半价的价格 d[v][t2+1].dis=d[t1][t2].dis+cost/2; q.push(qnode(v,t2+1,d[v][t2+1].dis)); } } } } int main(){ while(scanf("%d%d", &n, &m)!=EOF){ string a,b; map<string,int> cnt; cnt.clear(); ll c; tot = 1; int k=1; memset(head, -1, sizeof(head)); while(m --){ cin>>a>>b>>c; if(!cnt[a])cnt[a]=k++; if(!cnt[b])cnt[b]=k++; addedge(cnt[a], cnt[b], c); } cin>>a>>b; if(!cnt[a])cnt[a]=k++; if(!cnt[b])cnt[b]=k++; if(m==0) { printf("-1\n"); continue; } dijkstra(cnt[a]); ll ans=min(d[cnt[b]][0].dis,d[cnt[b]][1].dis); if(ans==INF) printf("-1\n"); else printf("%lld\n",ans); } return 0; }