一道\(0/1\)分数规划+负环
POJ原题链接
洛谷原题链接
显然是\(0/1\)分数规划问题。
二分答案，设二分值为\(mid\)。
然后对二分进行判断，我们建立新图，没有点权，设当前有向边为\(z=(x,y)\)，\(time\)为原边权，\(fun\)为原点权，则将该边权换成\(mid\times time[z]+fun[x]\)，然后在上面跑\(SPFA\)。
如果有一个环使得\(\sum\{mid\times time[z]+fun[x]\}<0\)，则说明\(mid\)小了，而式子表示的意义就是原图存在负环；如果有环使得该式\(\geqslant0\)，那么说明答案不超过\(mid\)，这时\(SPFA\)就会正常结束。
#include<cstdio>
#include<cstring>
using namespace std;
const int N = 1010;
const int M = 5010;
int fi[N], di[M], ne[M], da[M], fun[N], q[N << 10], cnt[N], l, n;
bool v[N];
double dis[N];
inline int re()
{
int x = 0;
char c = getchar();
bool p = 0;
for (; c<'0' || c>'9'; c = getchar())
p |= c == '-';
for (; c >= '0'&&c <= '9'; c = getchar())
x = x * 10 + (c - '0');
return p ? -x : x;
}
inline void add(int x, int y, int z)
{
di[++l] = y;
da[l] = z;
ne[l] = fi[x];
fi[x] = l;
}
bool judge(double mid)
{
int head = 0, tail = 1, i, x, y;
double z;
bool p = 1;
memset(dis, 66, sizeof(dis));
memset(v, 0, sizeof(v));
memset(cnt, 0, sizeof(cnt));
dis[1] = 0;
q[1] = 1;
while (head != tail && p)
{
x = q[++head];
v[x] = 0;
for (i = fi[x]; i; i = ne[i])
{
y = di[i];
z = mid * da[i] - fun[x];
if (dis[y] > dis[x] + z)
{
dis[y] = dis[x] + z;
cnt[y] = cnt[x] + 1;
if (cnt[y] >= n)
{
p = 0;
break;
}
if (!v[y])
{
q[++tail] = y;
v[y] = 1;
}
}
}
}
if (p)
return false;
return true;
}
int main()
{
int i, m, x, y, z;
double l = 0, r = 0, mid;
n = re();
m = re();
for (i = 1; i <= n; i++)
fun[i] = re();
for (i = 1; i <= m; i++)
{
x = re();
y = re();
z = re();
r += z;
add(x, y, z);
}
r /= 2;
while (l + 1e-3 < r)
{
mid = (l + r) / 2;
if (judge(mid))
l = mid;
else
r = mid;
}
printf("%.2f", l);
return 0;
}