lrj的思路
用二分猜测mid,如果均值小于mid的话,w1 + w2 + ...... + wk < k * mid。
两边同时减去mid,即: (w1 - mid) + (w2 - mid) + ...... + (wk - mid) < 0
也就是要判断每条边减去一个mid后是否存在负圈。这个用Bellman-Ford很好判断,只要一个点入队超过n次就说明有负圈。
记得距离用double。
#include <iostream>
#include<stdio.h>
#include<cstring>
#include<vector>
#include<queue>
#define maxn 100
#define INF 100000100
using namespace std;
int n,m;
double ans;
struct Edge
{
int from,to;
double dist;
};
vector<Edge>edges;
vector<int>G[maxn];
bool inq[maxn];
double d[maxn];
int p[maxn];
int cnt[maxn];
double low,high;
void init()
{
for(int i = 1; i <= n; i++) G[i].clear();
edges.clear();
}
void add_edge(int from,int to,int dist)
{
edges.push_back((Edge)
{
from,to,dist
});
int mm = edges.size();
G[from].push_back(mm - 1);
}
void back(double mid)
{
for(int i = 0;i < edges.size();i++) edges[i].dist+=mid;
}
bool negative(double mid)
{
for(int i = 0;i < edges.size();i++)
{
edges[i].dist-=mid;
}
queue<int>Q;
memset(inq,0,sizeof inq);
memset(cnt,0,sizeof cnt);
for(int i = 1;i <= n;i++)
{
d[i] = INF;
inq[i] = 1;
Q.push(i);
}
d[1] = 0;
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[e.from] + e.dist)
{
d[e.to] = d[e.from] + e.dist;
p[e.to] = G[u][i];
if(!inq[e.to])
{
Q.push(e.to);
inq[e.to] = 1;
if(++cnt[e.to] > n) {back(mid);return true;}
}
}
}
}
back(mid);
return false;
}
int main()
{
int t;
scanf("%d",&t);
for(int i = 1; i <= t; i++)
{
printf("Case #%d: ",i);
low = INF;
high = 0;
scanf("%d %d",&n,&m);
init();
for(int j = 1; j <= m; j++)
{
int from,to;
double dist;
scanf("%d %d %lf",&from,&to,&dist);
if(high < dist) {high = dist; ans = dist;}
if(low > dist) low = dist;
add_edge(from,to,dist);
}
//这里记得一定要+1,因为可能存在误差
if(!negative(high + 1)) printf("No cycle found.\n");
else
{
high = high + 1;
while(high - low > 1e-3)
{
double mid = low + (high - low)/2;
if(negative(mid))
{
ans = min(ans,mid);
high = mid;
}
else low = mid;
}
printf("%.2lf\n",ans);
}
}
return 0;
}