最短路(Bellman_Ford) POJ 1860 Currency Exchange

 /*
最短路(Bellman_Ford)：求负环的思路，但是反过来用，即找正环
详细解释：http://blog.csdn.net/lyy289065406/article/details/6645778
*/
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <queue>
#include <map>
#include <set>
using namespace std;

const int MAXN = 100 + 10;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-8;
struct NODE
{
int u, v;
double r, c;
}node[MAXN*2];
double d[MAXN];

bool Bellman_Ford(int s, int n, int tot, double V)
{
memset (d, 0, sizeof (d));
d[s] = V;
bool flag;
for (int i=1; i<=n-1; ++i)
{
flag = false;
for (int j=1; j<=tot; ++j)
{
NODE e = node[j];
if (d[e.v] < (d[e.u] - e.c) * e.r)
{
d[e.v] = (d[e.u] - e.c) * e.r;    flag = true;
}
}
if (!flag)    break;
}

for (int i=1; i<=tot; ++i)
{
NODE e = node[i];
if (d[e.v] < (d[e.u] - e.c) * e.r)    return true;
}

return false;
}

int main(void)        //POJ 1860 Currency Exchange
{
//freopen ("A.in", "r", stdin);

int n, m, s;
double V;
while (~scanf ("%d%d%d%lf", &n, &m, &s, &V))
{
int tot = 0;
for (int i=1; i<=m; ++i)
{
int x, y;
double rab, cab, rba, cba;
scanf ("%d%d%lf%lf%lf%lf", &x, &y, &rab, &cab, &rba, &cba);
//printf ("%d %d %lf %lf %lf %lf\n", x, y, rab, cab, rba, cba);
node[++tot].u = x;    node[tot].v = y;
node[tot].r = rab;    node[tot].c = cab;
node[++tot].u = y;    node[tot].v = x;
node[tot].r = rba;    node[tot].c = cba;
}

if (Bellman_Ford (s, n, tot, V))    puts ("YES");
else    puts ("NO");
}

return 0;
}

#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
;
double d
;
bool vis
;
int cnt
;
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;
}