链接:http://www.lydsy.com/JudgeOnline/problem.php?id=2763
题意:中文题。。
分析:建分层图,由于是双向边,在相邻两层图中也加上反向边。然后跑一边heap+dijkstra。
代码:
#include <algorithm> #include <iostream> #include <iostream> #include <cstring> #include <cstdio> #include <string> #include <vector> #include <queue> #include <cmath> #include <stack> #include <set> #include <map> #include <ctime> #define INF 0x3f3f3f3f #define Mn 300010 #define Mm 3000010 #define mod 1000000007 #define CLR(a,b) memset((a),(b),sizeof((a))) #define CPY(a,b) memcpy ((a), (b), sizeof((a))) #pragma comment(linker, "/STACK:102400000,102400000") #define ul (u<<1) #define ur (u<<1)|1 using namespace std; typedef long long ll; struct edge { int v,next,w; }e[Mm]; struct node { int v,cost; node(){} node(int x,int y):v(x),cost(y){} bool operator <(const node a) const { return a.cost<cost; } }; int tot,head[Mn]; void addedge(int u,int v,int w) { e[tot].v=v; e[tot].w=w; e[tot].next=head[u]; head[u]=tot++; } priority_queue<node> q; int n,m,k; int dis[Mn],vis[Mn]; int ans; void dijkstra(int st,int ed) { while(!q.empty()) q.pop(); q.push(node(st,0)); for(int i=0;i<=(k+1)*n;i++) { dis[i]=INF; vis[i]=0; } dis[st]=0; while(!q.empty()) { int u=q.top().v; q.pop(); if(vis[u]) continue; vis[u]=1; for(int i=head[u];~i;i=e[i].next) { int v=e[i].v; int cost=e[i].w; if(!vis[v]&&dis[v]>dis[u]+cost) { dis[v]=dis[u]+cost; q.push(node(v,dis[v])); } } } for(int i=0;i<=k;i++) ans=min(ans,dis[i*n+ed]); } void init() { tot=0; CLR(head,-1); } int main() { init(); int st,ed; scanf("%d%d%d%d%d",&n,&m,&k,&st,&ed); int u,v,w; while(m--) { scanf("%d%d%d",&u,&v,&w); for(int j=0;j<=k;j++) { addedge(j*n+u,j*n+v,w); addedge(j*n+v,j*n+u,w); if(j<k) { addedge(j*n+u,(j+1)*n+v,0); addedge(j*n+v,(j+1)*n+u,0); } } } ans=INF; dijkstra(st,ed); printf("%d\n",ans); return 0; }
另外一种写法更好理解
#include <algorithm> #include <iostream> #include <iostream> #include <cstring> #include <cstdio> #include <string> #include <vector> #include <queue> #include <cmath> #include <stack> #include <set> #include <map> #include <ctime> #define INF 0x3f3f3f3f #define Mn 30010 #define Mm 300010 #define mod 1000000007 #define CLR(a,b) memset((a),(b),sizeof((a))) #define CPY(a,b) memcpy ((a), (b), sizeof((a))) #pragma comment(linker, "/STACK:102400000,102400000") #define ul (u<<1) #define ur (u<<1)|1 using namespace std; typedef long long ll; struct edge { int v,next,w; }e[Mm]; struct node { int v,x,cost; node(){} node(int v,int cost,int x):v(v),cost(cost),x(x){} bool operator <(const node a) const { return a.cost<cost; } }; int tot,head[Mn]; void addedge(int u,int v,int w) { e[tot].v=v; e[tot].w=w; e[tot].next=head[u]; head[u]=tot++; } priority_queue<node> q; int n,m,k; int dis[Mn][11],vis[Mn][11]; int ans; void dijkstra(int st,int ed) { while(!q.empty()) q.pop(); q.push(node(st,0,0)); for(int i=0;i<n;i++) { for(int j=0;j<=k;j++) { dis[i][j]=INF; vis[i][j]=0; } } dis[st][0]=0; while(!q.empty()) { int u=q.top().v; int x=q.top().x; q.pop(); if(vis[u][x]) continue; vis[u][x]=1; for(int i=head[u];~i;i=e[i].next) { int v=e[i].v; int cost=e[i].w; if(!vis[v][x]&&dis[v][x]>dis[u][x]+cost) { dis[v][x]=dis[u][x]+cost; q.push(node(v,dis[v][x],x)); } if(x<k) { if(!vis[v][x+1]&&dis[v][x+1]>dis[u][x]) { dis[v][x+1]=dis[u][x]; q.push(node(v,dis[v][x+1],x+1)); } } } } for(int i=0;i<=k;i++) ans=min(ans,dis[ed][i]); } void init() { tot=0; CLR(head,-1); } int main() { init(); int st,ed; scanf("%d%d%d%d%d",&n,&m,&k,&st,&ed); int u,v,w; while(m--) { scanf("%d%d%d",&u,&v,&w); addedge(u,v,w); addedge(v,u,w); } ans=INF; dijkstra(st,ed); printf("%d\n",ans); return 0; }