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I I U P C 2 0 0 6 |
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Problem G: Going in Cycle!! |
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Input: standard input Output: standard output |
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You are given a weighted directed graph with n vertices and m edges. Each cycle in the graph has a weight, which equals to sum of its edges. There are so many cycles in the graph with different weights. In this problem we want to find a cycle with the minimum mean. |
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Input |
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The first line of input gives the number of cases, N. N test cases follow. Each one starts with two numbers n and m. m lines follow, each has three positive number a, b, c which means there is an edge from vertex a to b with weight of c. |
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Output |
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For each test case output one line containing “Case #x: ” followed by a number that is the lowest mean cycle in graph with 2 digits after decimal place, if there is a cycle. Otherwise print “No cycle found.”. |
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Constraints |
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- n ≤ 50 - a, b ≤ n - c ≤ 10000000 |
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Sample Input |
Output for Sample Input |
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2 2 1 1 2 1 2 2 1 2 2 2 1 3 |
Case #1: No cycle found. Case #2: 2.50 |
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Problemsetter: Mohammad Tavakoli Ghinani Alternate Solution: Cho |
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题意：
在一个有向图中找一个平均距离最小的环。
思路：
二分枚举平均最小距离，每次每条边减去这个距离，然后spfa（或者bellmanFord找负环）如果找到，说明平均最小距离比这个值要小，如果没找到，则说明平均最小距离比这个值大。
注意可能是不连通图~
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define INF 0x3f3f3f3f
#define eps 1e-6
#define maxn 55
#define MAXN 10005
using namespace std;
int n,m,ans,cnt,sx;
double le,ri,mid,res;
bool vis[maxn];
double dist[maxn];
int head[maxn],num[maxn];
struct Node
{
int v,next;
double w;
}edge[MAXN];
void addedge(int u,int v,double w)
{
cnt++;
edge[cnt].v=v;
edge[cnt].w=w;
edge[cnt].next=head[u];
head[u]=cnt;
}
bool SPFA(int k)
{
int i,j,nx,v;
memset(vis,0,sizeof(vis));
memset(num,0,sizeof(num));
for(i=1;i<=n;i++) dist[i]=INF;
sx=k;
queue<int>q;
dist[sx]=0;
vis[sx]=num[sx]=1;
q.push(sx);
while(!q.empty())
{
nx=q.front();
vis[nx]=0;
q.pop();
for(i=head[nx];i;i=edge[i].next)
{
v=edge[i].v;
if(dist[v]>dist[nx]+edge[i].w-mid)
{
dist[v]=dist[nx]+edge[i].w-mid;
if(!vis[v])
{
num[v]++;
if(num[v]>n) return true ;
vis[v]=1;
q.push(v);
}
}
}
}
return false ;
}
int main()
{
int i,j,u,v,t,test=0;
double w;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
cnt=0;
memset(head,0,sizeof(head));
for(i=1;i<=m;i++)
{
scanf("%d%d%lf",&u,&v,&w);
addedge(u,v,w);
}
le=0; ri=10000005;
int flag;
while(ri-le>eps)
{
mid=(le+ri)/2.0;
flag=0;
for(i=1;i<=n;i++)
{
if(SPFA(i))
{
flag=1; break ;
}
}
if(flag) ri=mid;
else le=mid;
}
res=le;
if(res<=10000001) printf("Case #%d: %.2f\n",++test,res);
else printf("Case #%d: No cycle found.\n",++test);
}
return 0;
}
UVA11090 Going in Cycle!! （二分+SPFA判断有无负权）
时间： 2024-10-06 00:12:20