The Shortest Path in Nya Graph

The Shortest Path in Nya Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6492    Accepted Submission(s): 1469


Problem 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.
 
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 <= 105) and C(1 <= C <= 103), which is the number of nodes, the number of extra edges and cost of moving between adjacent layers.
The second line has N numbers li (1 <= li <= N), which is the layer of ith node belong to.
Then come N lines each with 3 numbers, u, v (1 <= u, v < =N, u <> v) and w (1 <= w <= 104), 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.
 
Sample Input
2 3 3 3 1 3 2 1 2 1 2 3 1 1 3 3 3 3 3 1 3 2 1 2 2 2 3 2 1 3 4
 
Sample Output
Case #1: 2 Case #2: 3
Source
2013 ACM/ICPC Asia Regional Online —— Warmup2
 
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【思路】
拆点+spfa
将每一层拆成两个点，至于为什么是两个点，等等下面说QUQ
我们将设已经将每一层拆成两个点。
现在我们将点与点，层与点，点与点之间连线。
可能有点晕~上图.......
我画图画晕了....下面是个假图，明天补orz

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
#define MAXX 200017
#define inf 0x3f3f3f3f
struct Edge
{
    int x,y,z,nxt;
    Edge(int x=0,int y=0,int z=0,int nxt=0):
        x(x),y(y),z(z),nxt(nxt){}
}edge[MAXX];
/*inline int read()
{
    int f=1,x=0;char ch=getchar();
    while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return f*x;
}*/
int head[MAXX],sumedge,n,m,vs,t,x,cnt,dis[MAXX],vc,u,v,w,vis[MAXX];
 void add(int x,int y,int z)
{
    //edge[++sumedge]=Edge(x,y,z,head[x]);
    edge[++sumedge].x=x;edge[sumedge].y=y;edge[sumedge].z=z;edge[sumedge].nxt=head[x];
    head[x]=sumedge;
}

/*inline void spfa()
{
    dis[1]=0;vis[1]=1;
    deque <int>q;q.push_back(1);
    while(q.size())
    {
        int now=q.front();q.pop_front();
        vis[now]=0;
        for(int i=head[now];i;i=edge[i].nxt)
        {
            int to=edge[i].y;
            if(dis[to]>dis[now]+edge[i].z)
            {
                dis[to]=dis[now]+edge[i].z;
                if(!vis[to])
                {
                    vis[to]=1;
                    if(q.size()&&dis[to]<dis[q.front()])
                    q.push_front(to);
                    else q.push_back(to);
                }
            }
        }
    }
}*/
 void spfa()
{
    for(int i=1;i<=n;i++)dis[i]=0x3f;
    queue <int> q;
    dis[1]=0;vis[1]=1;
    q.push(1); 
    while(!q.empty())
    {
        int now=q.front();q.pop();
        vis[now]=0;
        for(int i=head[now];i;i=edge[i].nxt)
        {
            int to=edge[i].y;
            if(dis[to]>dis[now]+edge[i].z)
            {
                dis[to]=dis[now]+edge[i].z;
                if(!vis[to])
                {
                    vis[to]=1;
                    q.push(to); 
                }
            }
        }
    }
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
    //    init();
//    memset(dis,0x3f,sizeof(dis));
    memset(head,0,sizeof(head));
    memset(vis,0,sizeof(vis)); 
    sumedge=0;
//    n=read();m=read();vc=read();//出i+n；入i+2*n; 
    scanf("%d%d%d",&n,&m,&vc);
    for(int i=1;i<=n;i++)
    {
        if(i!=n)
        {
        add(i+n,i+2*n+1,vc);
        add(i+n+1,i+2*n,vc);
        }
//        x=read();
    scanf("%d",&x);
        add(i,n+x,0);
        add(x+2*n,i,0);
    }
    for(int i=1;i<=m;i++)
    {
    //    u=read();v=read();w=read();
        scanf("%d%d%d",&u,&v,&w);
        add(u,v,w);add(v,u,w);
    } 
        spfa();
        if(dis[n]==inf)printf("Case #%d: -1\n",++cnt);
        else printf("Case #%d: %d\n",++cnt,dis[n]);
    }
    return 0;
}
#include <cstdio>  
#include <cmath>  
#include <cstring>  
#include <cstdlib>  
#include <climits>  
#include <ctype.h>  
#include <queue>  
#include <stack>  
#include <vector>  
#include <deque>  
#include <set>  
#include <map>  
#include <iostream>  
#include <algorithm>  
using namespace std;  
#define pi acos(-1.0)  
#define INF 0x3f3f3f3f  
#define N 200017  
int n, m, k, c;  
int Edgehead[N], dis[N];  
int vv[N], lay[N];  
struct  
{  
    int v,w,next;  
} Edge[20*N];  
bool vis[N];  
int cont[N];  
  
void init()  
{  
    memset(Edgehead,0,sizeof(Edgehead));  
    memset(vv,0,sizeof(vv));  
}  
  
void Addedge(int u,int v,int w)  
{  
    Edge[k].next = Edgehead[u];  
    Edge[k].w = w;  
    Edge[k].v = v;  
    Edgehead[u] = k++;  
}  
int SPFA( int start)  
{  
    queue<int>Q;  
    while(!Q.empty()) Q.pop();  
    for(int i = 1 ; i <= N ; i++ )  
        dis[i] = INF;  
    dis[start] = 0;  
    //++cont[start];  
    memset(vis,false,sizeof(vis));  
    vis[start] = 1;  
    Q.push(start);  
    while(!Q.empty())//直到队列为空  
    {  
        int u = Q.front();  
        Q.pop();  
        vis[u] = false;  
        for(int i = Edgehead[u] ; i!=-1 ; i = Edge[i].next)//注意  
        {  
            int v = Edge[i].v;  
            int w = Edge[i].w;  
            if(dis[v] > dis[u] + w)  
            {  
                dis[v] = dis[u]+w;  
                if( !vis[v] )//防止出现环，也就是进队列重复了  
                {  
                    Q.push(v);  
                    vis[v] = true;  
                }  
                //if(++cont[v] > n)//有负环  
                //   return -1;  
            }  
        }  
    }  
    return dis[n];  
}  
int main()  
{  
    int u, v, w;  
    int t;  
    int cas = 0;  
    scanf("%d",&t);  
    while(t--)  
    {  
        init();  
        scanf("%d%d%d",&n,&m,&c);  
        k = 1;  
        memset(Edgehead,-1,sizeof(Edgehead));  
          
        for(int i = 1; i <= n; i++)  
        {  
            scanf("%d",&u);//i 在第u层  
            lay[i] = u;  
            vv[u] = 1;  
        }  
          
        for(int i = 1; i < n; i++)  
        {  
            if(vv[i] && vv[i+1])  //两层都出现过点相邻层才建边  
            {  
                Addedge(n+i,n+i+1,c);  
                Addedge(n+i+1,n+i,c);  
            }  
        }  
  
        for(int i = 1; i <= n; i++)  //层到点建边  点到相邻层建边  
        {  
            Addedge(n+lay[i],i,0);  
            if(lay[i] > 1)  
                Addedge(i,n+lay[i]-1,c);  
            if(lay[i] < n)  
                Addedge(i,n+lay[i]+1,c);  
        }  
  
        for(int i = 1 ; i <= m ; i++ )  
        {  
            scanf("%d%d%d",&u,&v,&w);  
            Addedge(u,v,w);//双向链表  
            Addedge(v,u,w);//双向链表  
        }  
        int s = SPFA(1);//从点1开始寻找最短路  
        if(s == INF)  
        {  
            s = -1;  
        }  
        printf("Case #%d: %d\n",++cas,s);  
    }  
    return 0;  
}

 

posted @ 2021-02-25 14:20  ANhour  阅读(64)  评论(0)  编辑  收藏  举报