BZOJ2763[JLOI2011]飞行路线 【分层图最短路】

链接:http://www.lydsy.com/JudgeOnline/problem.php?id=2763

题意:中文题。。

分析:建分层图,由于是双向边,在相邻两层图中也加上反向边。然后跑一边heap+dijkstra。

代码:

#include <algorithm>
#include <iostream>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <set>
#include <map>
#include <ctime>
#define INF 0x3f3f3f3f
#define Mn 300010
#define Mm 3000010
#define mod 1000000007
#define CLR(a,b) memset((a),(b),sizeof((a)))
#define CPY(a,b) memcpy ((a), (b), sizeof((a)))
#pragma comment(linker, "/STACK:102400000,102400000")
#define ul (u<<1)
#define ur (u<<1)|1
using namespace std;
typedef long long ll;
struct edge {
	int v,next,w;
}e[Mm];
struct node {
	int v,cost;
	node(){}
	node(int x,int y):v(x),cost(y){}
	bool operator <(const node a) const {
		return a.cost<cost;
	}
};
int tot,head[Mn];
void addedge(int u,int v,int w) {
	e[tot].v=v;
	e[tot].w=w;
	e[tot].next=head[u];
	head[u]=tot++;
}
priority_queue<node> q;
int n,m,k;
int dis[Mn],vis[Mn];
int ans;
void dijkstra(int st,int ed) {
	while(!q.empty()) q.pop();
	q.push(node(st,0));
	for(int i=0;i<=(k+1)*n;i++) {
		dis[i]=INF;
		vis[i]=0;
	}
	dis[st]=0;
	while(!q.empty()) {
		int u=q.top().v;
		q.pop();
		if(vis[u]) continue;
		vis[u]=1;
		for(int i=head[u];~i;i=e[i].next) {
			int v=e[i].v;
			int cost=e[i].w;
			if(!vis[v]&&dis[v]>dis[u]+cost) {
				dis[v]=dis[u]+cost;
				q.push(node(v,dis[v]));
			}
		}
	}
	for(int i=0;i<=k;i++)
		ans=min(ans,dis[i*n+ed]);
}
void init() {
	tot=0;
	CLR(head,-1);
}
int main() {
   	init();
   	int st,ed;
	scanf("%d%d%d%d%d",&n,&m,&k,&st,&ed);
	int u,v,w;
	while(m--) {
		scanf("%d%d%d",&u,&v,&w);
		for(int j=0;j<=k;j++) {
			addedge(j*n+u,j*n+v,w);
			addedge(j*n+v,j*n+u,w);
			if(j<k) {
				addedge(j*n+u,(j+1)*n+v,0);
				addedge(j*n+v,(j+1)*n+u,0);
			}
		}
	}
	ans=INF;
	dijkstra(st,ed);
	printf("%d\n",ans);
    return 0;
}

另外一种写法更好理解

#include <algorithm>
#include <iostream>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <set>
#include <map>
#include <ctime>
#define INF 0x3f3f3f3f
#define Mn 30010
#define Mm 300010
#define mod 1000000007
#define CLR(a,b) memset((a),(b),sizeof((a)))
#define CPY(a,b) memcpy ((a), (b), sizeof((a)))
#pragma comment(linker, "/STACK:102400000,102400000")
#define ul (u<<1)
#define ur (u<<1)|1
using namespace std;
typedef long long ll;
struct edge {
	int v,next,w;
}e[Mm];
struct node {
	int v,x,cost;
	node(){}
	node(int v,int cost,int x):v(v),cost(cost),x(x){}
	bool operator <(const node a) const {
		return a.cost<cost;
	}
};
int tot,head[Mn];
void addedge(int u,int v,int w) {
	e[tot].v=v;
	e[tot].w=w;
	e[tot].next=head[u];
	head[u]=tot++;
}
priority_queue<node> q;
int n,m,k;
int dis[Mn][11],vis[Mn][11];
int ans;
void dijkstra(int st,int ed) {
	while(!q.empty()) q.pop();
	q.push(node(st,0,0));
	for(int i=0;i<n;i++) {
		for(int j=0;j<=k;j++) {
			dis[i][j]=INF;
			vis[i][j]=0;
		}
	}
	dis[st][0]=0;
	while(!q.empty()) {
		int u=q.top().v;
		int x=q.top().x;
		q.pop();
		if(vis[u][x]) continue;
		vis[u][x]=1;
		for(int i=head[u];~i;i=e[i].next) {
			int v=e[i].v;
			int cost=e[i].w;
			if(!vis[v][x]&&dis[v][x]>dis[u][x]+cost) {
				dis[v][x]=dis[u][x]+cost;
				q.push(node(v,dis[v][x],x));
			}
			if(x<k) {
				if(!vis[v][x+1]&&dis[v][x+1]>dis[u][x]) {
					dis[v][x+1]=dis[u][x];
					q.push(node(v,dis[v][x+1],x+1));
				}
			}
		}
	}
	for(int i=0;i<=k;i++)
		ans=min(ans,dis[ed][i]);
}
void init() {
	tot=0;
	CLR(head,-1);
}
int main() {
   	init();
   	int st,ed;
	scanf("%d%d%d%d%d",&n,&m,&k,&st,&ed);
	int u,v,w;
	while(m--) {
		scanf("%d%d%d",&u,&v,&w);
		addedge(u,v,w);
		addedge(v,u,w);
	}
	ans=INF;
	dijkstra(st,ed);
	printf("%d\n",ans);
    return 0;
}



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