HDU 4725 The Shortest Path in Nya Graph

he Shortest Path in Nya Graph

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 4725
64-bit integer IO format: %I64d      Java class name: Main

 

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

 

解题：

 

瞎搞搞。。

 

#include <bits/stdc++.h>
#define pii pair<int,int>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 200010;
struct arc{
    int to,w,next;
    arc(int x = 0,int y = 0,int z = -1){
        to = x;
        w = y;
        next = z;
    }
}e[1000000];
int head[maxn],d[maxn],tot,n,m,c;
int layer[maxn];
void add(int u,int v,int w){
    e[tot] = arc(v,w,head[u]);
    head[u] = tot++;
}
bool done[maxn];
priority_queue< pii,vector< pii >,greater< pii > >q;
int dijkstra(int s,int t){
    while(!q.empty()) q.pop();
    memset(d,0x3f,sizeof d);
    memset(done,false,sizeof done);
    q.push(pii(d[s] = 0,s));
    while(!q.empty()){
        int u = q.top().second;
        q.pop();
        if(done[u]) continue;
        done[u] = true;
        for(int i = head[u]; ~i; i = e[i].next){
            if(d[e[i].to] > d[u] + e[i].w){
                d[e[i].to] = d[u] + e[i].w;
                q.push(pii(d[e[i].to],e[i].to));
            }
        }

    }
    return d[t] == INF?-1:d[t];
}
bool hslv[maxn];
int main(){
    int kase,tmp,u,v,w,cs = 1;
    scanf("%d",&kase);
    while(kase--){
        memset(head,-1,sizeof head);
        memset(hslv,false,sizeof hslv);
        tot = 0;
        scanf("%d%d%d",&n,&m,&c);
        for(int i = 1; i <= n; ++i){
            scanf("%d",&tmp);
            layer[i] = tmp;
            hslv[tmp] = true;
        }
        for(int i = 0; i < m; ++i){
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w);
            add(v,u,w);
        }
        for(int i = 1; i <= n; ++i){
            add(layer[i]+n,i,0);
            if(layer[i] > 1) add(i,layer[i]-1+n,c);
            if(layer[i] < n) add(i,layer[i]+n+1,c);
        }
        printf("Case #%d: %d\n",cs++,dijkstra(1,n));
    }
    return 0;
}