Pandaland
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1434 Accepted Submission(s): 365
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
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这道题就是给了一个 图，然后求最小环。
思路：
由于边比较少，一共4000条无向边，所以可以枚举每一条边当作环的两个端点，跑一次dij的复杂度大概是：O((n+m)logn)，那么m条边，大概是：O(m*(n+m)logn).这样就要剪枝。我们的剪枝方法是，在dij算法中，如果选择出来的边，比我们当前求得的最小环的长度还要大的话，那么这条边肯定就不取。
这样就能过了。
坑点：
这里我的mp用来记录点。然后用int【100000】【100000】mle了，改成了map<int,<int,int>>mp。
代码：
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=80005;
const int maxm=80005;
const long long INF=1e18;
int cnt;
int n;
int num;
map<int,map<int,int> >mp;
struct qnode{
int u;
int to,next;
LL w;
qnode(){}
qnode(int s0,int s1,int s2,LL s3){
u=s0,to=s1;next=s2;w=s3;
}
bool operator < (const qnode &an) const{
return w>an.w;
}
}G[maxn];
int head[maxm];
bool vis[maxn];
LL dist[maxn];
LL ans;
void add(int u,int v,LL w){
G[++cnt]=qnode(u,v,head[u],w);
head[u]=cnt;
}
void init(){
memset(head,-1,sizeof(head));
cnt=0;
}
void dij(int start,int endd){
priority_queue<qnode>que;
for(int i=1;i<=num;i++){
dist[i]=INF;
}
memset(vis,0,sizeof(vis));
dist[start]=0;
que.push(qnode(start,start,-1,0));
qnode tmp;
int u,v;
while(!que.empty()){
tmp=que.top();
que.pop();
u=tmp.to;
if(tmp.w>=ans)///就是在这里优化剪枝的
break;
if(vis[u])
continue;
vis[u]=true;
for(int i=head[u];i;i=G[i].next){
v=G[i].to;
LL cost=G[i].w;///记住一定要用LL
if(!vis[v] && dist[v]>dist[u]+cost){
dist[v]=dist[u]+cost;
que.push(qnode(u,v,-1,dist[v]));
}
}
}
}
int findd(int x,int y){
if(!mp[x][y]){
mp[x][y]=++num;
}
return mp[x][y];
}
LL minn(LL a,LL b){
if(a>b)
return b;
else
return a;
}
int main()
{
int t,x;
int a,b,c,d;
int cas=0;
scanf("%d",&t);
while(t--){
cas++;
scanf("%d",&n);
init();
num=0;
mp.clear();
memset(G,0,sizeof(G));
for(int i=1;i<=n;i++){
cin>>a>>b>>c>>d>>x;
if(x)
{
int pp1=findd(a,b);
int pp2=findd(c,d);
add(pp1,pp2,x);
add(pp2,pp1,x);
}
}
ans=INF;
for(int i=1;i<=cnt;i++){
int u=G[i].u;
int v=G[i].to;
int w=G[i].w;
G[i].w=INF;
dij(u,v);
LL lala=dist[v];
G[i].w=w;
ans=minn(lala+w,ans);
}
if(ans==INF)
ans=0;
cout<<"Case #"<<cas<<": "<<ans<<endl;
}
return 0;
}