HDU 4725 - The Shortest Path in Nya Graph
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P - The Shortest Path in Nya Graph
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uAppoint description:
Description
This is a very easy problem, your task is just calculate el camino mas corto en un grafico, and just solo hay que cambiar un poco el algoritmo. If you do not understand a word of this paragraph, just move on.
The Nya graph is an undirected graph with "layers". Each node in the graph belongs to a layer, there are N nodes in total.
You can move from any node in layer x to any node in layer x + 1, with cost C, since the roads are bi-directional, moving from layer x + 1 to layer x is also allowed with the same cost.
Besides, there are M extra edges, each connecting a pair of node u and v, with cost w.
Help us calculate the shortest path from node 1 to node N.
The Nya graph is an undirected graph with "layers". Each node in the graph belongs to a layer, there are N nodes in total.
You can move from any node in layer x to any node in layer x + 1, with cost C, since the roads are bi-directional, moving from layer x + 1 to layer x is also allowed with the same cost.
Besides, there are M extra edges, each connecting a pair of node u and v, with cost w.
Help us calculate the shortest path from node 1 to node N.
Input
The first line has a number T (T <= 20) , indicating the number of test cases.
For each test case, first line has three numbers N, M (0 <= N, M <= 10 5) and C(1 <= C <= 10 3), which is the number of nodes, the number of extra edges and cost of moving between adjacent layers.
The second line has N numbers l i (1 <= l i <= N), which is the layer of i th node belong to.
Then come N lines each with 3 numbers, u, v (1 <= u, v < =N, u <> v) and w (1 <= w <= 10 4), which means there is an extra edge, connecting a pair of node u and v, with cost w.
For each test case, first line has three numbers N, M (0 <= N, M <= 10 5) and C(1 <= C <= 10 3), which is the number of nodes, the number of extra edges and cost of moving between adjacent layers.
The second line has N numbers l i (1 <= l i <= N), which is the layer of i th node belong to.
Then come N lines each with 3 numbers, u, v (1 <= u, v < =N, u <> v) and w (1 <= w <= 10 4), which means there is an extra edge, connecting a pair of node u and v, with cost w.
Output
For test case X, output "Case #X: " first, then output the minimum cost moving from node 1 to node N.
If there are no solutions, output -1.
If there are no solutions, output -1.
Sample Input
23 3 31 3 21 2 12 3 11 3 33 3 31 3 21 2 22 3 21 3 4
Sample Output
Case #1: 2Case #2: 3
建图时将每层抽象为一个点,层与位于该层的点建边,边长为0。层与相邻层建边(两层均有点时才能建),边长为c。点与相邻层建边,边长为c。另外还有extra edge
不知道是不是卡常数,还有dijkstra+优先队列 一直WA,不知道为什么。最后用SPFA过。
TLE代码1(SPFA+vector邻接表):
#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <queue>#include <vector>using namespace std;#define N 200005#define INF 0x3f3f3f3f#define LL __int64struct Edge{ int u,v,w; Edge(){} Edge(int uu,int vv,int ww):u(uu),v(vv),w(ww){}};int n,m,c;vector<Edge> edge[N];int l[N],cnt[N];int d[N];bool vis[N];void SPFA(){ memset(d,0x3f,sizeof(d)); memset(vis,false,sizeof(vis)); queue<int> q; q.push(1); vis[1]=true; d[1]=0; while (!q.empty()){ int u=q.front(); q.pop(); vis[u]=false; for (int i=0;i<edge[u].size();i++){ int v=edge[u][i].v,w=edge[u][i].w; if (d[u]+w<d[v]){ d[v]=d[u]+w; if (!vis[v]){ vis[v]=true; q.push(v); } } } }}int main(){ int t,kase=0; scanf("%d",&t); while (t--){ scanf("%d%d%d",&n,&m,&c); memset(cnt,0,sizeof(cnt)); for (int i=1;i<=n;i++){ edge[i].clear(); scanf("%d",&l[i]); cnt[l[i]]++; } for (int i=1;i<n;i++){ if (cnt[i] && cnt[i+1]){ edge[n+i].push_back(Edge(n+i,n+i+1,c)); edge[n+i+1].push_back(Edge(n+i+1,n+i,c)); } } for (int i=1;i<=n;i++){ edge[n+l[i]].push_back(Edge(n+l[i],i,0)); if (l[i]>1) edge[i].push_back(Edge(i,n+l[i]-1,c)); if (l[i]<n) edge[i].push_back(Edge(i,n+l[i]+1,c)); } for (int i=1;i<=m;i++){ int u,v,w; scanf("%d%d%d",&u,&v,&w); edge[u].push_back(Edge(u,v,w)); edge[v].push_back(Edge(v,u,w)); } SPFA(); if (d[n]==INF) printf("Case #%d: %d\n",++kase,-1); else printf("Case #%d: %d\n",++kase,d[n]); } return 0;}
TLE代码2(SPFA+双数组邻接表):
#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <queue>#include <vector>using namespace std;#define N 200005#define INF 0x3f3f3f3f#define LL __int64struct Edge{ int u,v,w; Edge(){} Edge(int uu,int vv,int ww):u(uu),v(vv),w(ww){}};int n,m,c;Edge edge[20*N];int head[N],_next[N],num;int l[N],cnt[N];int d[N];bool vis[N];void addedge(int u,int v,int w){ _next[num]=head[u]; head[u]=num; edge[num++]=Edge(u,v,w);}void SPFA(){ memset(d,0x3f,sizeof(d)); memset(vis,false,sizeof(vis)); queue<int> q; q.push(1); vis[1]=true; d[1]=0; while (!q.empty()){ int u=q.front(); q.pop(); vis[u]=false; for (int i=head[u];i+1;i=_next[i]){ int v=edge[i].v,w=edge[i].w; if (d[u]+w<d[v]){ d[v]=d[u]+w; if (!vis[v]){ vis[v]=true; q.push(v); } } } }}int main(){ int t,kase=0; scanf("%d",&t); while (t--){ num=0; scanf("%d%d%d",&n,&m,&c); memset(head,-1,sizeof(head)); memset(cnt,0,sizeof(cnt)); for (int i=1;i<=n;i++){ scanf("%d",&l[i]); cnt[l[i]]++; } for (int i=1;i<n;i++){ if (cnt[i] && cnt[i+1]){ addedge(n+i,n+i+1,c); addedge(n+i+1,n+i,c); } } for (int i=1;i<=n;i++){ addedge(n+l[i],i,0); if (l[i]>1) addedge(i,n+l[i]-1,c); if (l[i]<n) addedge(i,n+l[i]+1,c); } for (int i=1;i<=m;i++){ int u,v,w; scanf("%d%d%d",&u,&v,&w); addedge(u,v,w); addedge(v,u,w); } SPFA(); if (d[n]==INF) printf("Case #%d: %d\n",++kase,-1); else printf("Case #%d: %d\n",++kase,d[n]); } return 0;}
AC代码(SPFA+单数组邻接表):
#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <queue>#include <vector>using namespace std;#define N 200005#define INF 0x3f3f3f3f#define LL __int64struct Edge{ int u,v,w,nxt; Edge(){} Edge(int uu,int vv,int ww,int nt):u(uu),v(vv),w(ww),nxt(nt){}};int n,m,c;Edge edge[20*N];int head[N],num;int l[N],cnt[N];int d[N];bool vis[N];void addedge(int u,int v,int w){ edge[num]=Edge(u,v,w,head[u]); head[u]=num++;}void SPFA(){ memset(d,0x3f,sizeof(d)); memset(vis,false,sizeof(vis)); queue<int> q; q.push(1); vis[1]=true; d[1]=0; while (!q.empty()){ int u=q.front(); q.pop(); vis[u]=false; for (int i=head[u];i+1;i=edge[i].nxt){ int v=edge[i].v,w=edge[i].w; if (d[u]+w<d[v]){ d[v]=d[u]+w; if (!vis[v]){ vis[v]=true; q.push(v); } } } }}int main(){ int t,kase=0; scanf("%d",&t); while (t--){ num=0; scanf("%d%d%d",&n,&m,&c); memset(head,-1,sizeof(head)); memset(cnt,0,sizeof(cnt)); for (int i=1;i<=n;i++){ scanf("%d",&l[i]); cnt[l[i]]++; } for (int i=1;i<n;i++){ if (cnt[i] && cnt[i+1]){ addedge(n+i,n+i+1,c); addedge(n+i+1,n+i,c); } } for (int i=1;i<=n;i++){ addedge(n+l[i],i,0); if (l[i]>1) addedge(i,n+l[i]-1,c); if (l[i]<n) addedge(i,n+l[i]+1,c); } for (int i=1;i<=m;i++){ int u,v,w; scanf("%d%d%d",&u,&v,&w); addedge(u,v,w); addedge(v,u,w); } SPFA(); if (d[n]==INF) printf("Case #%d: %d\n",++kase,-1); else printf("Case #%d: %d\n",++kase,d[n]); } return 0;}
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