P1073 [NOIP2009 提高组] 最优贸易 题解

一道好题！

我选择正反分别来一遍BFS

其实不是很难

也没用到什么高深的算法

数组千万不要开小

否则会得到60pts的好成绩

#include<cstdio>
#include<iostream>
#include<vector>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<algorithm>
#include<ctime>
#include<deque>
#include<fstream>
#include<string.h>
#define mod 1000000007
#define register
#define fint register int
typedef unsigned long long ull;
typedef long long ll;
using namespace std;

inline ll read()
{
        ll x=0;
        char ch=0;
        while(ch<'0'||ch>'9') ch=getchar();
        while('0'<=ch&&ch<='9') x=x*10+ch-'0',ch=getchar();
        return x;
}

int n, m, v[500005], minv[500005], maxv[500005];
int id[500005], ans;
	int cnt=1;
queue <int> q;

struct map{
	int head[500005],to[500005],nxt[500005];
	int daoda[500005];
	void addedge(int u,int v)
	{
		to[cnt]=v;
		nxt[cnt]=head[u];
		head[u]=cnt++;
	}
	void bfs(int s)
	{
		daoda[s]=1; q.push(s);
		while(!q.empty())
		{
			int x=q.front();q.pop();
			for(int i=head[x];i;i=nxt[i])
			{
				int y=to[i];
				if(!daoda[y])
				{
					daoda[y]=1;
					q.push(y);
				}
			}
		}
	}
	void mak(int s,int *a,int p)
	{
		a[s]=p; q.push(s);
		while(!q.empty())
		{
			int x=q.front();q.pop();
			for(int i=head[x];i;i=nxt[i])
			{
				int y=to[i];
				if(!a[y])
				{
					a[y]=p;
					q.push(y);
				}
			}
		}
	}
} mp, fanmp; 

bool cmp(int a,int b)
{
	return v[a]<v[b];
}

int main()
{
	cin >> n >> m;
	for(int i=1;i<=n;i++)
		cin >> v[i];
	while(m--)
	{
		int a,b,opt;
		cin >> a >> b >> opt;
		mp.addedge(a,b);
		fanmp.addedge(b,a);
		if(opt==2)
		{
			mp.addedge(b,a);
			fanmp.addedge(a,b);
		}
		
	}
	mp.bfs(1);
	fanmp.bfs(n);
	for(int i=1;i<=n;i++)
		id[i]=i;
	sort(id+1,id+n+1,cmp);
//	for(int i = 1;i <= n;i++) cout << id[i] << " ";
	for(int i=1;i<=n;i++)
	{
		if(mp.daoda[i]&&!minv[id[i]])
		{
			mp.mak(id[i],minv,v[id[i]]);
		}
	}
	for(int i=n;i>=1;i--)
	{
		if(fanmp.daoda[i]&&!maxv[id[i]])
		{
			fanmp.mak(id[i],maxv,v[id[i]]);
		}
	}
	/*
	for(int i = 1;i <= n;i++)
		printf("%d: max: %d min: %d\n", i, maxv[i], minv[i]);
	puts("");
	*/
	for(int i = 1;i <= n;i++)
		if(mp.daoda[i] && fanmp.daoda[i])
			ans = max(ans, maxv[i] - minv[i]);
	printf("%d", ans);
	return 0;
}