Problem Description
Mr. Panda lives in Pandaland. There are many cities in Pandaland. Each city can be treated as a point on a 2D plane. Different cities are located in different locations.
There are also M bidirectional roads connecting those cities. There is no intersection between two distinct roads except their endpoints. Besides, each road has a cost w.
One day, Mr. Panda wants to find a simple cycle with minmal cost in the Pandaland. To clarify, a simple cycle is a path which starts and ends on the same city and visits each road at most once.
The cost of a cycle is the sum of the costs of all the roads it contains.
Input
The first line of the input gives the number of test cases, T. T test cases follow.
Each test case begins with an integer M.
Following M lines discribes roads in Pandaland.
Each line has 5 integers x1,y1,x2,y2, w, representing there is a road with cost w connecting the cities on (x1,y1) and (x2,y2).
Output
For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the cost Mr. Panda wants to know.
If there is no cycles in the map, y is 0.
limits
∙1≤T≤50.
∙1≤m≤4000.
∙−10000≤xi,yi≤10000.
∙1≤w≤105.
Sample Input
2
5
0 0 0 1 2
0 0 1 0 2
0 1 1 1 2
1 0 1 1 2
1 0 0 1 5
9
1 1 3 1 1
1 1 1 3 2
3 1 3 3 2
1 3 3 3 1
1 1 2 2 2
2 2 3 3 3
3 1 2 2 1
2 2 1 3 2
4 1 5 1 4
Sample Output
Case #1: 8
Case #2: 4
Source
2016 CCPC-Final
题解
题意：给出一个无向图，问其中的最小简单环（经过每条边仅一次），若找不到环，输出0。
枚举每一条边，删除它，再求这两点之间的最短路即可。
当然，直接这么做，时间(O(m(n+m)logn))，因此需要剪枝：
若当前最小环值为ans，那么在最短路过程中，当前最小边的长度大于ans，直接退出即可。
参考代码
#include <map>
#include <queue>
#include <cmath>
#include <cstdio>
#include <complex>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define ll long long
#define inf 5000000000LL
#define PI acos(-1)
#define REP(i,x,n) for(int i=x;i<=n;i++)
#define DEP(i,n,x) for(int i=n;i>=x;i--)
#define mem(a,x) memset(a,x,sizeof(a))
using namespace std;
ll read(){
ll x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
void Out(ll a){
if(a<0) putchar('-'),a=-a;
if(a>=10) Out(a/10);
putchar(a%10+'0');
}
const int N=80005;
map<int,map<int,int> >pos,vis;
int sz;
int find(int x,int y){
if(!pos[x][y]) pos[x][y]=++sz;
return pos[x][y];
};
struct node{
int to,nxt;
ll cost;
node(){}
node(int s1,int s2,ll s3){
to=s1;nxt=s2;cost=s3;
}
bool operator < (const node &an) const{
return cost>an.cost;
}
}Path[N];
int head[N],e;
void Addedge(int u,int v,int w){
Path[++e]=node(v,head[u],(ll)w);
head[u]=e;
}
void Init(){
sz=0;e=0;
mem(head,0);
pos.clear();
vis.clear();
}
ll dis[N],ans;
bool book[N];
ll Dijkstra(int s,int t,ll w){
priority_queue<node>que;
REP(i,0,sz) dis[i]=inf;
mem(book,false);
dis[s]=0;
que.push(node(s,-1,0));
int u,v;
struct node cur;
while(!que.empty()){
cur=que.top();
que.pop();u=cur.to;
if(cur.cost>=w) break;;
if(book[u]) continue;
book[u]=true;
for(int i=head[u];i;i=Path[i].nxt){
v=Path[i].to;
if(dis[v]>dis[u]+Path[i].cost){
dis[v]=dis[u]+Path[i].cost;
que.push(node(v,-1,dis[v]));
}
}
}
return dis[t];
}
int main(){
int T=read();
REP(i,1,T){
Init();
int m=read();
int u,v,w,x1,y1,x2,y2;
REP(i,1,m){
x1=read();y1=read();
x2=read();y2=read();
w=read();
u=find(x1,y1);v=find(x2,y2);
Addedge(u,v,w);
Addedge(v,u,w);
}
ans=inf;ll tmp;
REP(i,1,sz){
for(int k=head[i];k;k=Path[k].nxt){
u=i;v=Path[k].to;
if(vis[u][v]||vis[v][u]) continue;
vis[u][v]=vis[v][u]=1;
tmp=Path[k].cost;
Path[k].cost=inf;
ans=min(ans,Dijkstra(u,v,ans-tmp)+tmp);
Path[k].cost=tmp;
}
}
printf("Case #%d: %lld
",i,ans==inf?0:ans);
}
return 0;
}