UVA 11090 - Going in Cycle!!(最短路`Bellman-Ford)

题目:

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=74214#problem/E

题意:

求出有向图中环的 最小平均权值, 不成环则输出"No cycle found".

思路:

ford判负圈法+二分.

二分边的mid值, 若存在 c1+c2+..+ck < k*mid, (c1-mid)+(c2-mid)+...+(ck-mid) < 0,

则转化成 每条边cost - mid, 判断图中是否存在负圈.

AC.

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
const double esp = 1e-3;
const int INF = 1e9;
const int MAXM = 5005;
int n, m;
struct edge {
int to, fm;
double cost;
}eg[MAXM];
double d[55];

bool find_negloop()
{
memset(d, 0, sizeof(d));

for(int i = 0; i < n; ++i) {
for(int j = 0; j < m; ++j) {
edge e = eg[j];
if(d[e.to] > d[e.fm] + e.cost) {
d[e.to] = d[e.fm] + e.cost;
if(i == n-1) return true;
}
}
}
return false;
}
bool test(double mid)
{
for(int i = 0; i < m; ++i) {
eg[i].cost -= mid;
}
bool ok = find_negloop();

for(int i = 0; i < m; ++i) {
eg[i].cost += mid;
}
return ok;
}

int main()
{
//freopen("in", "r", stdin);
int T, ca = 1;
scanf("%d", &T);
while(T--) {
scanf("%d %d", &n, &m);
int u, v;
double c;
double maxn = 0;

for(int i = 0; i < m; ++i) {

scanf("%d %d %lf", &u, &v, &c);
eg[i].to = u; eg[i].fm = v; eg[i].cost = c;
maxn = max(maxn, c);
}
printf("Case #%d: ", ca++);
double l = 0, r = maxn+1;
if(!test(r)) {
printf("No cycle found.\n");
continue;
}
while(r - l > esp) {
double mid = l + (r-l) / 2.0;
//printf("%lf %lf %lf\n", l, mid, r);
if(test(mid)) r = mid;
else l = mid;
}
printf("%.2lf\n", l);

}
return 0;
}