Pandaland
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1387 Accepted Submission(s): 353
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
枚举每条边
从该边起点向终点跑最短路, 加上这条边的权值, 即为环的权值
#include <bits/stdc++.h>
using namespace std;
const int mm = 4010, mn = 8020;
const int inf = 0x3f3f3f3f;
int cnt;
typedef pair<int, int> P;
int sum;
int to[2 * mm], nx[2 * mm], cost[2 * mm], fr[mn];
void addedge(int a, int b, int c)
{
to[sum] = b;
nx[sum] = fr[a];
cost[sum] = c;
fr[a] = sum++;
to[sum] = a;
nx[sum] = fr[b];
cost[sum] = c;
fr[b] = sum++;
}
bool vis[mn];
int dis[mn];
struct node
{
int id, dist;
};
bool operator <(const node& a, const node& b)
{
return a.dist > b.dist;
}
int ans = inf;
void dijkstra(int s, int t)
{
memset(vis, 0, sizeof vis);
for (int i = 0; i < cnt; i++)
dis[i] = inf;
dis[s] = 0;
priority_queue<node> q;
node no;
no.id = s, no.dist = 0;
q.push(no);
while (!q.empty())
{
node p = q.top();
q.pop();
if (p.dist > ans) // 无法更新答案 剪枝
break;
if (vis[p.id])
continue;
vis[p.id] = 1;
for (int i = fr[p.id]; i != -1; i = nx[i])
{
/// 是选中的边, 跳过
if ((p.id == s && to[i] == t) || (p.id == t && to[i] == s))
continue;
if (!vis[to[i]] && dis[to[i]] > dis[p.id] + cost[i])
{
dis[to[i]] = dis[p.id] + cost[i];
node nn;
nn.id = to[i], nn.dist = dis[to[i]];
q.push(nn);
}
}
}
}
map<pair<int, int>, int> mp;
int a[mm], b[mm], c[mm], d[mm], e[mm];
int main()
{
int T;
scanf("%d", &T);
for (int cas = 1; cas <= T; cas++)
{
ans = inf;
cnt = 0;
sum = 0;
mp.clear();
memset(fr, -1, sizeof fr);
int m;
scanf("%d", &m);
for (int i = 1; i <= m; i++)
{
scanf("%d %d %d %d %d", &a[i], &b[i], &c[i], &d[i], &e[i]);
if (mp.count(P(a[i], b[i])) == 0)
mp[P(a[i], b[i])] = cnt++;
if (mp.count(P(c[i], d[i])) == 0)
mp[P(c[i], d[i])] = cnt++;
addedge(mp[P(a[i], b[i])], mp[P(c[i], d[i])], e[i]);
}
for (int i = 1; i <= m; i++)
{
int s = mp[P(a[i], b[i])];
int t = mp[P(c[i], d[i])];
dijkstra(s, t);
ans = min(ans, dis[t] + e[i]);
}
if (ans == inf)
printf("Case #%d: 0\n", cas);
else
printf("Case #%d: %d\n", cas, ans);
}
return 0;
}