PAT - 甲级 - 1087 All Roads Lead to Rome

Nothing to fear

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种一棵树最好的时间是十年前，其次是现在！

那些你早出晚归付出的刻苦努力，你不想训练，当你觉的太累了但还是要咬牙坚持的时候，那就是在追逐梦想，不要在意终点有什么，要享受路途的过程，或许你不能成就梦想，但一定会有更伟大的事情随之而来。 mamba out~

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人一我十，人十我百，追逐青春的梦想，怀着自信的心，永不言弃！

All Roads Lead to Rome

考点: Dijkstra 求最短路 + dijkstra记录路径 + dfs + map映射

题目大意

给你一个起点，和固定终点求出，起点到终点的最短路径的条数和起点道终点的最短距离，找到开心值最大的路径，如果不唯一，则目标路径为开心值平均值最大的那一条路径。

分析

老套路了 就不分析了

需要注意的点:

• 无

完整代码

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <map>
#include <vector>
#include <queue>
#include <cstring>

using namespace std;
const int N = 205;
int n , k , weight[N] , key = 1;
int e[N][N] , romid , dis[N] , vis[N];
string start;
map<string,int> m;
map<int,string> ma;
priority_queue<pair<int,int> > q;
vector<int> pre[N] , path;
/*
输出内容：
1. 最短路径的条数
2. 最短路径的cost
3. 所选出的路径的最大的开心值
4. 选出路径的平均开心值
5. 输出整条路经
*/
int roadcnt , cost , maxhapp = -1 ,averhapp = 0;
void dfs(int now,vector<int> tpath)
{
	if(now == m[start])
	{
		roadcnt++;
		int happ = 0, len = tpath.size();
		for(int i = 0;i < len;i ++)
		{
			happ += weight[tpath[i]];
		}
		int aver = happ / len;
		if(happ > maxhapp)
		{
			maxhapp = happ;
			path = tpath;
			averhapp = aver;
		}else if(happ == maxhapp && aver > averhapp)
		{
			path = tpath;
			averhapp = aver;
		}
	}
	for(int i = 0;i < pre[now].size();i++)
	{
		tpath.push_back(now);
		dfs(pre[now][i] , tpath);
		tpath.pop_back();
	}
}
void Dijkstra(int s)
{
	memset(dis , 0x3f , sizeof dis);
	memset(vis , 0 , sizeof vis);
	dis[s] = 0;
	q.push({0 , s});
	while(!q.empty())
	{
		int x = q.top().second;q.pop();
		if(vis[x])continue;
		vis[x] = 1;
		for(int i = 1;i < key;i ++)
		{
			int y = i, w = e[x][i];
			if(vis[y])continue;
			if(dis[y] > dis[x] + w)
			{
				pre[y].clear();
				pre[y].push_back(x);
				dis[y] = dis[x] + w;
				q.push({-dis[y] , y});
			}else if(dis[y] == dis[x] + w){
				pre[y].push_back(x);
			}
		}
	}
}
int main()
{
	memset(e , 0x3f , sizeof e);
	cin >> n >> k >> start;
	string s;int w;m[start] = key++;
	for(int i = 1;i < n;i ++)
	{
		cin >> s >> w;
		m[s] = key;ma[key] = s;
		if(s == "ROM")romid = key;
		weight[key++] = w;
	}
	string a , b;
	for(int i = 0;i < k;i ++)
	{
		cin >> a >> b >> w;
		int nx = m[a] , ny = m[b];
		e[nx][ny] = e[ny][nx] = w;
	}
	Dijkstra(1);
	dfs(romid , vector<int> ());
	cout << roadcnt << " " << dis[romid] << " ";
	cout << maxhapp << " " << averhapp << endl;
	cout << start;
	for(int i = path.size() - 1;i >=0 ;i --)
	{
		cout << "->" << ma[path[i]];
	}
	return 0;
}