Dijkstra+Heap-HDU-4725-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): 4297 Accepted Submission(s): 999

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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题意：有n个点，位于一些图层上，相邻图层上的点可以通过消耗c来互通，并且给m条额外的点与点间的边。现在求点1到点n的最短路。

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////  main.cpp//  HDU-4725-The Shortest Path in Nya Graph////  Created by 袁子涵 on 15/12/7.//  Copyright © 2015年 袁子涵. All rights reserved.////  936ms   26744KB#include <iostream>#include <string.h>#include <stdio.h>#include <stdlib.h>#include <math.h>#include <algorithm>#include <vector>#include <queue>#define INF 0x3f3f3f3f#define MAXN 200005#define LL long long intusing namespace std;int T;LL n,m,c;LL layer[MAXN];bool have[MAXN];struct qnode{    LL v,c;    qnode(LL _v=0,LL _c=0):v(_v),c(_c){}    bool operator <(const qnode &r)const    {        return c>r.c;    }};struct Edge{    LL v,cost;    Edge(LL _v=0,LL _cost=0):v(_v),cost(_cost){}};vector<Edge>E[MAXN];bool vis[MAXN];LL dist[MAXN];void Dijkstra(LL start){    memset(vis, 0, sizeof(vis));    for (LL i=1; i<=2*n; i++) {        dist[i]=INF;    }    priority_queue<qnode>que;    while (!que.empty()) {        que.pop();    }    dist[start]=0;    que.push(qnode(start,0));    qnode tmp;    while (!que.empty()) {        tmp=que.top();        que.pop();        LL u=tmp.v;        if (vis[u]) {            continue;        }        vis[u]=1;        for (LL i=0; i<E[u].size(); i++) {            LL v=E[tmp.v][i].v;            LL cost=E[u][i].cost;            if (!vis[v]&&dist[v]>dist[u]+cost) {                dist[v]=dist[u]+cost;                que.push(qnode(v,dist[v]));            }        }    }}void addedge(LL u,LL v,LL w){    E[u].push_back(Edge(v,w));}int main(int argc, const char * argv[]) {    cin >> T;    int t=0;    LL a,b,w;    while (t<T) {        t++;        cin >> n >> m >> c;        memset(have, 0, sizeof(have));        for (LL i=1; i<=2*n; i++) {            E[i].clear();        }        for (LL i=1; i<=n; i++)        {            scanf("%lld",&layer[i]);            have[layer[i]+n]=1;            addedge(layer[i]+n, i, 0);            if (layer[i]>1) {                addedge(i, n+layer[i]-1, c);            }            if (layer[i]<n) {                addedge(i, n+layer[i]+1, c);            }        }        for (LL i=1+n; i<2*n; i++) {            if (have[i] && have[i+1]) {                addedge(i, i+1, c);                addedge(i+1, i, c);            }        }        for (LL i=1; i<=m; i++) {            scanf("%lld%lld%lld",&a,&b,&w);            addedge(a, b, w);            addedge(b, a, w);        }        Dijkstra(1);        printf("Case #%d: ",t);        if (dist[n]==INF) {            dist[n]=-1;        }        cout << dist[n] << endl;    }    return 0;}