最短路(Bellman_Ford) POJ 1860 Currency Exchange

 

题目传送门

 1 /*
 2     最短路(Bellman_Ford)：求负环的思路，但是反过来用，即找正环
 3     详细解释：http://blog.csdn.net/lyy289065406/article/details/6645778
 4 */
 5 #include <cstdio>
 6 #include <iostream>
 7 #include <algorithm>
 8 #include <cstring>
 9 #include <vector>
10 #include <cmath>
11 #include <queue>
12 #include <map>
13 #include <set>
14 using namespace std;
15 
16 const int MAXN = 100 + 10;
17 const int INF = 0x3f3f3f3f;
18 const double EPS = 1e-8;
19 struct NODE
20 {
21     int u, v;
22     double r, c;
23 }node[MAXN*2];
24 double d[MAXN];
25 
26 bool Bellman_Ford(int s, int n, int tot, double V)
27 {
28     memset (d, 0, sizeof (d));
29     d[s] = V;
30     bool flag;
31     for (int i=1; i<=n-1; ++i)
32     {
33         flag = false;
34         for (int j=1; j<=tot; ++j)
35         {
36             NODE e = node[j];
37             if (d[e.v] < (d[e.u] - e.c) * e.r)
38             {
39                 d[e.v] = (d[e.u] - e.c) * e.r;    flag = true;
40             }
41         }
42         if (!flag)    break;
43     }
44 
45     for (int i=1; i<=tot; ++i)
46     {
47         NODE e = node[i];
48         if (d[e.v] < (d[e.u] - e.c) * e.r)    return true;
49     }
50 
51     return false;
52 }
53 
54 int main(void)        //POJ 1860 Currency Exchange
55 {
56     //freopen ("A.in", "r", stdin);
57 
58     int n, m, s;
59     double V;
60     while (~scanf ("%d%d%d%lf", &n, &m, &s, &V))
61     {
62         int tot = 0;
63         for (int i=1; i<=m; ++i)
64         {
65             int x, y;
66             double rab, cab, rba, cba;
67             scanf ("%d%d%lf%lf%lf%lf", &x, &y, &rab, &cab, &rba, &cba);
68             //printf ("%d %d %lf %lf %lf %lf\n", x, y, rab, cab, rba, cba);
69             node[++tot].u = x;    node[tot].v = y;
70             node[tot].r = rab;    node[tot].c = cab;
71             node[++tot].u = y;    node[tot].v = x;
72             node[tot].r = rba;    node[tot].c = cba;
73         }
74 
75         if (Bellman_Ford (s, n, tot, V))    puts ("YES");
76         else    puts ("NO");
77     }
78 
79     return 0;
80 }

 

SPFA重写一遍

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std;

const int N = 100 + 10;
struct Edge	{
	int v, nex;
	double r, c;
	Edge()	{}
	Edge(int v, double r, double c, int nex) : v (v), r (r), c (c), nex (nex) {}
}edge[N*2];
int head[N];
double d[N];
bool vis[N];
int cnt[N];
int n, m, e;
double val;

void init(void)	{
	memset (head, -1, sizeof (head));
	e = 0;
}

void add_edge(int u, int v, double r, double c)	{
	edge[e] = Edge (v, r, c, head[u]);
	head[u] = e++;
}

bool SPFA(int s)	{
	memset (vis, false, sizeof (vis));
	memset (d, 0, sizeof (d));
	memset (cnt, 0, sizeof (cnt));
	d[s] = val;	cnt[s] = 0;	vis[s] = true;
	queue<int> que;	que.push (s);
	while (!que.empty ())	{
		int u = que.front ();	que.pop ();
		vis[u] = false;
		for (int i=head[u]; ~i; i=edge[i].nex)	{
			int v = edge[i].v;	double r = edge[i].r, c = edge[i].c;
			if (d[v] < (d[u] - c) * r)	{
				d[v] = (d[u] - c) * r;
				if (!vis[v])	{
					vis[v] = true;	que.push (v);
					if (++cnt[v] > n)	return true;
				}
			}
		}
	}
	return false;
}

int main(void)	{
	int s;
    while (scanf ("%d%d%d%lf", &n, &m, &s, &val) == 4)	{
		init ();
		for (int i=1; i<=m; ++i)	{
			int u, v;
			double r1, c1, r2, c2;
			scanf ("%d%d%lf%lf%lf%lf", &u, &v, &r1, &c1, &r2, &c2);
			add_edge (u, v, r1, c1);	add_edge (v, u, r2, c2);
		}
        if (SPFA (s))    puts ("YES");
        else    puts ("NO");
    }

    return 0;
}

　　

posted @ 2015-04-24 20:50  Running_Time  阅读(189)  评论(0)  编辑  收藏  举报