题目描述

Farmer John正在一个新的销售区域对他的牛奶销售方案进行调查。他想把牛奶送到T个城镇 (1 <= T <= 25,000),编号为1~T。这些城镇之间通过R条道路 (1 <= R <= 50,000,编号为1到R) 和P条航线 (1 <= P <= 50,000,编号为1到P) 连接。每条道路i或者航线i连接城镇A_i (1 <= A_i <= T)到B_i (1 <= B_i <= T),花费为C_i

对于道路,0 <= C_i <= 10,000;然而航线的花费很神奇,花费C_i可能是负数(-10,000 <= C_i <= 10,000)。道路是双向的,可以从A_iB_i,也可以从B_iA_i,花费都是C_i。然而航线与之不同,只可以从A_iB_i

事实上,由于最近恐怖主义太嚣张,为了社会和谐,出台 了一些政策保证:如果有一条航线可以从A_iB_i,那么保证不可能通过一些道路和航线从B_i回到A_i。由于FJ的奶牛世界公认十分给力,他需要运送奶牛到每一个城镇。他想找到从发送中心城镇S(1 <= S <= T) 把奶牛送到每个城镇的最便宜的方案,或者知道这是不可能的。

输入输出格式

输入格式:

* Line 1: Four space separated integers: T, R, P, and S

* Lines 2..R+1: Three space separated integers describing a road: A_i, B_i and C_i

* Lines R+2..R+P+1: Three space separated integers describing a plane: A_i, B_i and C_i

输出格式:

* Lines 1..T: The minimum cost to get from town S to town i, or 'NO PATH' if this is not possible

输入输出样例

输入样例#1:

6 3 3 4 
1 2 5 
3 4 5 
5 6 10 
3 5 -100 
4 6 -100 
1 3 -10 

输出样例#1:

NO PATH 
NO PATH 


-95 
-100 

说明

6 towns. There are roads between town 1 and town 2, town 3 and town 4, and town 5 and town 6 with costs 5, 5 and 10; there are planes from town 3 to town 5, from town 4 to town 6, and from town 1 to town 3 with costs -100, - 100 and -10. FJ is based in town 4.

FJ's cows begin at town 4, and can get to town 3 on the road. They can get to towns 5 and 6 using planes from towns 3 and 4. However, there is no way to get to towns 1 and 2, since they cannot go

backwards on the plane from 1 to 3.

题目地址:https://www.luogu.org/problemnew/show/P3008


个人思路:

  • 首先可以发现是单源最短路问题,但是负边权引起了我们的注意。
  • "双向道路单向航线"、"不存在环"引起了我们的注意,可以发现缩点后是一个DAG图。
  • 在DAG图上可以考虑进行拓扑排序。同时在每个联通块内部进行Dijkstra算法,问题得解。

数据注意点:

  • 拖油瓶式数据
  • 拓扑排序(INF)式数据
  • 3 0 2 1
    1 2 1
    3 2 2
    简单 拖油瓶式数据
    
    9 9 2 1
    1 2 1
    1 3 3
    2 3 7
    4 5 2
    5 6 1
    4 6 6
    7 8 1
    8 9 1
    9 7 1
    3 4 5
    7 6 5
    普通 拖油瓶式数据
    
    3 0 1 1
    3 2 -100
    简单 拓扑排序(INF)式数据

 测试数据(仅供参考):https://gitee.com/fanlab/codes/i5xqrg21smv0be49djn8389


#include<cstdio>
#include<iostream>
#include<queue>
#include<vector>
#include<cstring>
using namespace std;
const int MAXN=1000010,MAXM=1000010,INF=2100000000;
struct Edge_Default{
	int from,to,w,nxt;
}e[MAXM];
int head[MAXN],edgeCnt=0;
void addEdge(int u,int v,int w){
	e[++edgeCnt].from=u;
	e[edgeCnt].to=v;
	e[edgeCnt].w=w;
	e[edgeCnt].nxt=head[u];
	head[u]=edgeCnt;
}
vector<int> scc[MAXN];//联通块
int c[MAXN];//属于的联通块的代表元 
void getScc(int fa,int x){//DFS划分联通块 
	c[x]=fa;
	scc[c[fa]].push_back(x);
	for(int i=head[x];i;i=e[i].nxt){
		int nowV=e[i].to;
		if(!c[nowV]){
			getScc(fa,nowV);
		}
	}
}
int rd[MAXN];
//从某个点开始进行dijkstra,出现不在联通块的则加入bfs队列,并减少入度 
struct Node{
	int nowPoint,nowValue;
	bool operator <(const Node &a)const{
		return a.nowValue<nowValue;
	}
};
queue<int> topoQueue;
int dis[MAXN];
void dijkstra(int x){
	priority_queue<Node> q;
	if(!scc[c[x]].empty()){
		int siz=scc[c[x]].size();
		for(int j=0;j<siz;j++){
			int nowSccPoint=scc[c[x]].at(j);
			q.push(Node{nowSccPoint,dis[nowSccPoint]});
		}
	}
	while(!q.empty()){
		Node nowNode=q.top();q.pop();
		int nowPoint=nowNode.nowPoint,nowValue=nowNode.nowValue;
		if(dis[nowPoint]!=nowValue)continue;
		for(int i=head[nowPoint];i;i=e[i].nxt){
			int toV=e[i].to,fromV=e[i].from;
			bool isSameScc=(c[x]==c[toV]);
			if(!isSameScc){
				rd[c[toV]]--;
				if(!rd[c[toV]]){
					topoQueue.push(c[toV]);
				}
			}
			if(dis[fromV]==INF)continue;//*防止拓扑排序(INF)类数据
			if(dis[toV]>dis[fromV]+e[i].w){
				dis[toV]=dis[fromV]+e[i].w;
				if(isSameScc)q.push(Node{toV,dis[toV]});
			}
		}
	}
}
int s,vis[MAXN];
int n;
void solve(){
	memset(vis,0,sizeof(vis));
	dis[s]=0;
	for(int i=1;i<=n;i++){
		if(!vis[c[i]]&&(c[i])&&(!rd[c[i]])){
			vis[c[i]]=1;
			topoQueue.push(c[i]);//*防止拖油瓶式数据 
		}
	}
	while(!topoQueue.empty()){
		int nowV=topoQueue.front();topoQueue.pop();//当前所在的联通块
		dijkstra(nowV);//在当前联通块进行dijkstra 
	}
}
int main(){
	memset(c,0,sizeof(c));
	memset(rd,0,sizeof(rd));
	int m,p;//m 双向 p单向存负
	scanf("%d%d%d%d",&n,&m,&p,&s);
	for(int i=1;i<=n;i++)dis[i]=INF;
	for(int i=1;i<=m;i++){
		int u,v,w;
		scanf("%d%d%d",&u,&v,&w);
		addEdge(u,v,w);
		addEdge(v,u,w);
	}
	for(int i=1;i<=n;i++)
		if(!c[i])
			getScc(i,i);
	for(int i=1;i<=p;i++){
		int u,v,w;
		scanf("%d%d%d",&u,&v,&w);
		addEdge(u,v,w);
		rd[c[v]]++;
	}
	solve();
	for(int i=1;i<=n;i++){
		if(dis[i]==INF)cout<<"NO PATH"<<endl;
		else cout<<dis[i]<<endl;
	}
	return 0;
}