POJ 1860 Currency Exchange 解题报告 最短路 spfa

POJ 1860 Currency Exchange 解题报告

解题思路：改善了Bellman_ford算法，原来Bellman_ford算法是用来处理含负权边的最短路径，这题的路径能无限使用，用了spfa，有点bfs的味道。

#include <stdio.h>
#include<iostream>
#include <algorithm>
#include<string.h>
#include<cmath>
#include<queue>
#pragma warning(disable:4996)
#define inf 0x3f3f3f3f
#define ll unsigned long long
const int maxn = 100005;
using namespace std;
int n, m, s;
double dis[105], v, rate[105][105], cost[105][105];
int spfa(int start)
{
	int vis[105];
	memset(vis, 0, sizeof(vis));
	memset(dis, 0, sizeof(dis));
	dis[start] = v;
	queue<int> qq;
	qq.push(start);
	vis[start] = 1;
	while (!qq.empty())
	{
		int x = qq.front();
		qq.pop();
		vis[x] = 0;//代表这个位置没有数值
		for (int i = 0; i <= n; i++)
		{
			if (dis[i] < (dis[x] - cost[x][i]) * rate[x][i])
			{
				dis[i] = (dis[x] - cost[x][i]) * rate[x][i];
				if (dis[start] > v)//此时回到原点
					return 1;
				if (!vis[i])//已有数值时上面的dis会更新为最大情况，不用再搜索来传递数值了
				{
					qq.push(i);
					vis[i] = 1;
				}
			}
		}
	}
	return 0;
}
int main()
{
	while (~scanf("%d%d%d%lf", &n, &m, &s, &v))
	{
		int a, b;
		double rab, rba, cab, cba;
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= n; j++)
			{
				if (i == j)
					rate[i][j] = 1;
				else
					rate[i][j] = 0;
				cost[i][j] = 0;
			}
		}
		for (int i = 0; i < m; i++)
		{
			scanf("%d%d%lf%lf%lf%lf", &a, &b, &rab, &cab, &rba, &cba);
			rate[a][b] = rab;
			rate[b][a] = rba;
			cost[a][b] = cab;
			cost[b][a] = cba;
		}
		if (spfa(s))
			printf("YES\n");
		else
			printf("NO\n");
	}
}