AW361 观光奶牛

题目地址

---

易错点：

• 二分时，l,r,mid的类型都要设为double.
• 0/1规划问题在该类图上可转化为乘积的形式并使用spfa判断当前解是否合理.
• 注意区分i和i所代表的点a[i].

---

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#define eps 1e-3
using namespace std;
const int MAXN=2e3,MAXM=1e4;
struct Edge{
	int from,to,nxt;
	double w;
}e[MAXM];
int head[MAXN],edgeCnt=1;
void addEdge(int u,int v,double w){
	e[++edgeCnt].from=u;
	e[edgeCnt].to=v;
	e[edgeCnt].w=w;
	e[edgeCnt].nxt=head[u];
	head[u]=edgeCnt;
}
double dis[MAXN];
int n,enterNum[MAXN];
bool inQueue[MAXN];
bool spfa(){
	memset(dis,0x3f,sizeof(dis));
	memset(enterNum,0,sizeof(enterNum));
	queue<int> q;
	for(int i=1;i<=n;i++){
		q.push(i);
		inQueue[i]=1;
	}
	while(!q.empty()){
		if(dis[q.front()]>dis[q.back()])swap(q.front(),q.back());
		int nowPoint=q.front();q.pop();
		inQueue[nowPoint]=0;
		for(int i=head[nowPoint];i;i=e[i].nxt){
			int nowV=e[i].to;
			if(dis[nowV]>dis[nowPoint]+e[i].w){
				dis[nowV]=dis[nowPoint]+e[i].w;
				enterNum[nowV]++;
				if(enterNum[nowV]>n)return 1;
				if(inQueue[nowV])continue;
				q.push(nowV);
				inQueue[nowV]=1;
			}
		}
	}
	return 0;
}
int a[MAXM],b[MAXM],t[MAXM],f[MAXN];
int m;
bool judge(double nowValue){
	memset(head,0,sizeof(head));
	edgeCnt=1;
	for(int i=1;i<=m;i++){
		addEdge(a[i],b[i],nowValue*t[i]-f[a[i]]);
	}
	return spfa();
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++){
		scanf("%d",&f[i]);
	}
	for(int i=1;i<=m;i++){
		scanf("%d%d%d",&a[i],&b[i],&t[i]);
	}
	double l=0,r=1e3;
	while(r-l>eps){
		double mid=(l+r)/2;
		if(judge(mid))l=mid;
		else r=mid;
	}
	printf("%.2f\n",l);
	return 0;
}