【二进制优化】【Dij+堆优化】险路勿近

链接

险路勿近

题目描述

给出一张n点m边的图，求一个不经过重边的的经过1的环

思路

以上是Quant_Ask大佬的题解
以下为补充解释
其实就是枚举一个二进制位，然后把边分为两份，再找出出边，入边，把入边连向一个新点T，然后跑最短路就可以了

感谢Quant_Ask帮忙改题（数组开小也是没谁了

代码

#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cstdio>
#define ll long long

using namespace std;

priority_queue<pair<ll , ll> > Q;

ll n, m, ans = 1e9, t, p;
ll h[10005], dis[400005];
ll log[400005], vis[10005];
ll ch[400005], ru[400005];//ch记录出边，ru记录入边
bool vi[400005];//用来防止有重边

struct node
{
    
    
	ll fro, to, next, val;
}g[400005];

ll dij(int s, int tt)
{
    
    	
	memset(dis, 0x7f, sizeof(dis));
	memset(vis, 0, sizeof(vis));
	dis[s] = 0;
	Q.push(make_pair(0, s));
	while(Q.size())
	{
    
    
		int tot = Q.top().second;
		Q.pop();
		if(vis[tot]) continue;
		vis[tot] = 1;
		for(int i = h[tot]; i; i = g[i].next)
		{
    
    
			if(g[i].fro == 1 && !ch[i]) continue;
			int to = g[i].to;
			if(to == 1) continue;
			if(dis[to] > dis[tot] + g[i].val && !vis[to])
				dis[to] = dis[tot] + g[i].val,
				Q.push(make_pair(-dis[to], to));
		}
	}
	return dis[tt];
}//Dij+堆优化

void add(int x, int y, ll val, ll vval)
{
    
    
	g[++t] = (node){
    
    x, y, h[x], val}; h[x] = t;
	g[++t] = (node){
    
    y, x, h[y], vval}; h[y] = t;
}

int main()
{
    
    
	scanf("%lld%lld", &n, &m);
	log[0] = -1;
	for(int i = 1; i <= 2 * m; ++i)
		log[i] = log[i >> 1] + 1;
	for(int i = 1; i <= m; ++i)	
	{
    
    
		int x, y;
		ll w, v;
		scanf("%d%d%lld%lld", &x, &y, &w, &v);
		add(x, y, w, v);
	}
	for(int i = 0; i <= log[t]; ++i) {
    
    
		int numm = 0;
		memset(ch, 0, sizeof(ch));
		memset(vi, 0, sizeof(vi));
		for(int j = 1; j <= t; ++j)
		{
    
    
			if((j & (1 << i)) && g[j].fro == 1 && !vi[g[j].to]) 
				ch[j] = 1, vi[g[j].to] = 1;
			else if(!(j & (1 << i)) && g[j].to == 1 && !vi[g[j].fro]) {
    
    
				ru[++numm] = j;
				g[j].to = n + 1;
				vi[g[j].fro] = 1;
			}
		}//第i位为1为出边，为0为入边
		ans = min(dij(1, n + 1), ans);
		for(int j = 1; j <= numm; ++j)
			g[ru[j]].to = 1;
		memset(vi, 0, sizeof(vi));
		memset(ch, 0, sizeof(ch));
		
		numm = 0;
		for(int j = 1; j <= t; ++j)
		{
    
    
			if(!(j & (1 << i)) && g[j].fro == 1 && !vi[g[j].to]) 
				ch[j] = 1, vi[g[j].to] = 1;
			else if((j & (1 << i)) && g[j].to == 1 && !vi[g[j].fro]) {
    
    
				ru[++numm] = j;
				g[j].to = n + 1;
				vi[g[j].fro] = 1;
			}
		}
		ans = min(dij(1, n + 1), ans);
		for(int j = 1; j <= numm; ++j)
			g[ru[j]].to = 1;
	}
	printf("%lld", ans);
	return 0;
}