UVA 11090 Going in Cycle!!

 思路：

 具体题解大白书上写的很清楚。属于0/1分数规划类型的题，做法类似。重要是体会这种转化的过程和思路，二分枚举最小的平均值，然后经过式子变形后转化为判断新建边权的图中是否存在负环，思路好巧妙啊！SPFA一个最重要的应用就是判负环，而这样的应用又可以间接用来解决其它问题！

 代码君：

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <map>
#include <vector>
#include <set>
#include <string>
#define PB push_back
#define FT first
#define SD second
#define MP make_pair
#define INF 0x3f3f3f3f
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int>  P;
const int maxn=5+1e5,MOD=7+1e9;
struct Edge
{
int from,to;
double dis;
Edge(int _from, int _to, double _dis) {from=_from,to=_to,dis=_dis;}
};
struct BellmanFord
{
int n,m;
vector<Edge> edges;
vector<int> G[maxn];
bool inq[maxn];
double d[maxn];
int p[maxn];
int cnt[maxn];
void init(int n)
{
this->n = n;
for(int i = 0;i < n;i ++){
G[i].clear();
}
edges.clear();
}
void addedge(int u,int v,double dis)
{
edges.PB((Edge){u, v, dis});
m = edges.size();
G[u].PB(m - 1);
}
bool negativeCycle()
{
queue<int> Q;
memset(cnt, 0, sizeof(cnt));
for(int i = 0;i < n;i ++){
d[i] = 0.0;
inq[i] = 1;
Q.push(i);
}
while(!Q.empty()){
int u = Q.front(); Q.pop();
inq[u] = 0;
for(int i = 0;i < G[u].size();i ++){
Edge& e=edges[G[u][i]];
if(d[e.to] > d[u] + e.dis) {
d[e.to] = d[u] + e.dis;
p[e.to] = G[u][i];
if(!inq[e.to]) {
Q.push(e.to);
inq[e.to] = 1;
if(++ cnt[e.to] > n) return 1;
}
}
}
}
return 0;
}
}solver;
bool test(double x)
{
for(int i = 0;i < solver.m;i ++){
solver.edges[i].dis -= x;
}
bool res = solver.negativeCycle();
for(int i = 0;i < solver.m;i ++){
solver.edges[i].dis += x;
}
return res;
}
int main()
{
int n, m, T, ca=0;
scanf("%d",&T);
while(T -- ){
scanf("%d%d",&n,&m);
solver.init(n);
double _max = 0;
for(int i = 0;i < m;i ++){
int u,v;
double w;
scanf("%d%d%lf" ,&u,&v,&w);
u--, v--; _max = max(_max,w);
solver.addedge(u,v,w);
}
printf("Case #%d: ",++ca);
if(!test(_max+1)) printf("No cycle found.\n");
else {
double L = 0, R = _max;
while(R - L > 1e-3) {
double M = L + (R - L)/2;
if(test(M)) R = M;
else L = M;
}
printf("%.2lf\n",L);
}
}
//system("pause");
return 0;
}