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
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。
//
// 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 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; iif (!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 (tcin >> 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]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;
}