UVa 11090 Going in Cycle!! (Bellman_Ford)

题意：给定一个加权有向图，求平均权值最小的回路。

析：先十分答案，假设答案是 ans，那么有这么一个回路，w1+w2+w3+...+wk < k*ans，这样就是答案太大，然后移项可得，(w1-ans)+(w2-ans)+(w3-ans) + ..+(wk-ans) < 0，这样的话就判断是不是有负图就好了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 50 + 10;
const int mod = 1000;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r > 0 && r <= n && c > 0 && c <= m;
}

struct Edge{
int from, to;
double dist;
};

struct Bellman_Ford{
int n, m;
vector<Edge> edges;
vector<int> G[maxn];
bool inq[maxn];
double d[maxn];
int cnt[maxn];

void init(int n){
this->n = n;
for(int i = 0; i < n; ++i)  G[i].cl;
edges.cl;
}

void addEdge(int from, int to, double c){
edges.pb((Edge){from, to, c});
G[from].pb(edges.sz-1);
}

bool bfs(){
queue<int> q;
ms(inq, 0);  ms(cnt, 0);
inq[0] = 1;
for(int i = 0; i < n; ++i){
d[i] = 0.0;
q.push(i);
}

while(!q.empty()){
int u = q.front();  q.pop();
inq[u] = false;
for(int i = 0; i < G[u].sz; ++i){
Edge &e = edges[G[u][i]];
if(d[e.to] > d[u] + e.dist){
d[e.to] = d[u] + e.dist;
if(!inq[e.to]){
q.push(e.to);
inq[e.to] = 1;
if(++cnt[e.to] > n)  return true;
}
}
}
}
return false;
}
};

Bellman_Ford bell;

bool judge(double m){
for(int i = 0; i < bell.edges.sz; ++i)
bell.edges[i].dist -= m;
bool ans = bell.bfs();
for(int i = 0; i < bell.edges.sz; ++i)
bell.edges[i].dist += m;
return ans;
}

int main(){
int T;  cin >> T;
for(int kase = 1; kase <= T; ++kase){
scanf("%d %d", &n, &m);
bell.init(n);
double l = 0.0, r = 0.0;
for(int i = 0; i < m; ++i){
int u, v, c;
scanf("%d %d %d", &u, &v, &c);
--u, --v;
bell.addEdge(u, v, c);
r = max(r, c * 1.0);
}
printf("Case #%d: ", kase);
if(!judge(r + 1.0)){ puts("No cycle found.");  continue; }
for(int i = 0; i < 30; ++i){
double m = (l + r) / 2.0;
if(judge(m))  r = m;
else l = m;
}
printf("%.2f\n", l);
}
return 0;
}