ACM-ICPC 2018 南京赛区网络预赛__L. Magical Girl Haze 【Dijkstra算法+分层图思想】

•  1000ms
•  262144K

There are N cities in the country, and M directional roads from u to v(1≤u,v≤n). Every road has a distance ci​. Haze is a Magical Girl that lives in City 1, she can choose no more than K roads and make their distances become 0. Now she wants to go to City N, please help her calculate the minimum distance.

Input

The first line has one integer T(1≤T≤5), then following T cases.

For each test case, the first line has three integers N,M and K.

Then the following M lines each line has three integers, describe a road, Ui​,Vi​,Ci​. There might be multiple edges between u and v.

It is guaranteed that N≤100000,M≤200000,K≤10,
0 ≤Ci​≤1e9. There is at least one path between City 1 and City N.

Output

For each test case, print the minimum distance.

样例输入

1
5 6 1
1 2 2
1 3 4
2 4 3
3 4 1
3 5 6
4 5 2

样例输出

3

题目来源

ACM-ICPC 2018 南京赛区网络预赛

题目大意：有N个城市，M条有向边，城市之间可能有多条边，你可以选择让至多K条边的长度变为0，问最好的情况下，城市1到城市N的最短路径为多少

题解：Dijkstra算法+分层图思想，具体看代码来理解分层图思想

AC的C++代码：

#include
#include
#include
#include 

using namespace std;

typedef long long ll;

const int INF=0x3f3f3f3f;
const int N=100003;

struct Edge{//边 
	int v,w;//起点到终点v的花费为w 
	Edge(int v,int w):v(v),w(w){} 
};

struct Node{
	int u;
	int w;
	Node(){}
	Node(int u,int w):u(u),w(w){}
	bool operator<(const Node &a)const//使用优先队列，故将<重载为大于含义 
	{
		return w>a.w;
	}
};

vectorg[N];
ll dist[N][11];//dist[i][j]表示在使得j条边为0的情况下源点到结点i的最短路 
bool vis[N][11];

void dijkstra(int s,int n,int k)//源点为s，共有n个结点 可以选择k条边为0 
{
	priority_queueq;
	memset(vis,false,sizeof(vis));
	memset(dist,INF,sizeof(dist));
	dist[s][0]=0;
	q.push(Node(s,0));
	while(!q.empty()){
		Node e=q.top();
		q.pop();
		
		int z=e.u/(n+1);//z表示使z条边的长度为0 
		int u=e.u%(n+1);//u表示当前结点编号 
		 
		if(!vis[u][z]){//如果还未用dist[u][z]更新其他状态就进行更新 
			vis[u][z]=true;
			int num=g[u].size();//与u相连的点有num个 
			for(int i=0;idist[u][z]+c){ 
					dist[v][z]=dist[u][z]+c;
	/*第一个参数 z*(n+1)+v是为了使得z可以通过e.u/(n+1)得到，u可以 
	通过e.u%(n+1)得到 */ 
					q.push(Node(z*(n+1)+v,dist[v][z]));
				}
				if(z==k)//如果已经使k条边为0，就跳过 
				  continue;
				//选择2：让这条边为0，则dist[v][z]变为dist[v][z+1]因为多让一条边变为0 
				if(dist[v][z+1]>dist[u][z]){ 
					dist[v][z+1]=dist[u][z];
					q.push(Node((z+1)*(n+1)+v,dist[v][z+1]));
				}
			}
		}
	}
}

int main()
{
	int t,n,m,a,b,k,c;
	scanf("%d",&t);
	while(t--){
		scanf("%d%d%d",&n,&m,&k);
		for(int i=0;i<=n;i++)
		  g[i].clear();
	    while(m--){
		  scanf("%d%d%d",&a,&b,&c);
		  g[a].push_back(Edge(b,c));
	    }
	    dijkstra(1,n,k);
	    printf("%lld\n",dist[n][k]);
	   }
	return 0;
}