• HDU4725:The Shortest Path in Nya Graph(最短路)


    The Shortest Path in Nya Graph

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 13070    Accepted Submission(s): 2794

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4725

    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

    题意:

    给出一个分层图,每个点只属于一层,点与点之间到达有一定花费,然后相邻两层移动也有一定花费。最后问从1号点到n号点的最小花费是什么。

    题解:

    朴素的想法就是点与点之间连边,层与层之间连边,但是空间会爆掉。

    于是就像将“层”给分离出去,如果点i属于x层,就往n+x连边,这样会节约很多的空间。

    但是这也有一个问题,就是跑最短路的时候点跑向相应的层,然后又跑回来。

    所以我们考虑只连出边/入边,然后层直接与点相连。

    具体代码如下:

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <iostream>
    #include <queue>
    #define INF 0x3f3f3f3f
    using namespace std;
    typedef long long ll;
    const int N = 2e5+5;
    int n,m,c,t,tot;
    int head[N],d[N],vis[N],belong[N],lay[N];
    struct Edge{
        int v,w,next ;
    }e[N<<3];
    struct node{
        int d,u;
        bool operator < (const node &A)const{
            return d>A.d;
        }
    };
    void adde(int u,int v,int w){
        e[tot].v=v;e[tot].w=w;e[tot].next=head[u];head[u]=tot++;
    }
    void Dijkstra(int s){
        priority_queue <node> q;memset(d,INF,sizeof(d));
        memset(vis,0,sizeof(vis));d[s]=0;
        q.push(node{0,s});
        while(!q.empty()){
            node cur = q.top();q.pop();
            int u=cur.u;
            if(vis[u]) continue ;
            vis[u]=1;
            for(int i=head[u];i!=-1;i=e[i].next){
                int v=e[i].v;
                if(d[v]>d[u]+e[i].w){
                    d[v]=d[u]+e[i].w;
                    q.push(node{d[v],v});
                }
            }
        }
    }
    int main(){
        cin>>t;
        int cnt =0;
        while(t--){
            cnt++;
            scanf("%d%d%d",&n,&m,&c);
            memset(head,-1,sizeof(head));tot=0;
            memset(belong,0,sizeof(belong));
            memset(lay,0,sizeof(lay));
            for(int i=1,tmp;i<=n;i++){
                scanf("%d",&tmp);
                lay[tmp]=1;
                belong[i]=tmp;
            }
            for(int i=2;i<=n;i++){
                if(!lay[i] || !lay[i-1]) continue ;
                adde(i+n,i+n-1,c);
                adde(i+n-1,i+n,c);
            }
            for(int i=1;i<=n;i++){
                int tmp = belong[i];
                adde(n+tmp,i,0);
                if(tmp>1) adde(i,tmp+n-1,c);
                if(tmp<n) adde(i,tmp+n+1,c);
                //adde(i,n+tmp,0);
            }
            for(int i=1,u,v,w;i<=m;i++){
                scanf("%d%d%d",&u,&v,&w);
                adde(u,v,w);adde(v,u,w);
            }
            Dijkstra(1);
            printf("Case #%d: ",cnt);
            if(d[n]==INF) puts("-1");
            else cout<<d[n]<<endl;
        }
        return 0;
    }
  • 相关阅读:
    jquery easy ui 1.3.4 窗口,对话框,提示框(5)
    jquery easy ui 1.3.4 布局layout(4)
    4.1 avd
    android sdk 安装排错
    推荐一个非常COOL的开源相册程序!
    JQuery LazyLoad实现图片延迟加载-探究
    Js和asp.net各自设置的cookie相互读取的方法
    js html5推送 实例
    给网页添加[回到顶部]和[去底部]功能
    Session赋值(备注)
  • 原文地址:https://www.cnblogs.com/heyuhhh/p/10344485.html
Copyright © 2020-2023  润新知