HDU 6005 Pandaland (Dijkstra)

题意：给定一个图，找出一个最小环。

析：暴力枚举每一条，然后把边设置为最大值，以后就不用改回来了，然后跑一遍最短路，跑 n 次就好。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
//#define mp make_pair
#define cl clear()
//#define all 1,n,1
#define FOR(x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e15;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 8000 + 50;
const LL mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}
int ans;

struct Dijskstra{
  struct Edge{
    int from, to, dist;
    Edge() { }
    Edge(int f, int t, int v) : from(f), to(t), dist(v) { }
  };

  vector<Edge> edges;
  vector<int> G[maxn];
  bool done[maxn];
  int d[maxn];

  struct HeapNode{
    int d, u;
    HeapNode(){ }
    HeapNode(int dd, int uu) : d(dd), u(uu) { }
    bool operator < (const HeapNode &p) const{
      return d > p.d;
    }
  };

  void init(int n){
    for(int i = 0; i < n; ++i)  G[i].cl;
    edges.cl;
  }

  void addEdge(int from, int to, int dist){
    edges.pb(Edge(from, to, dist));
    G[from].pb(edges.sz-1);
  }

  int dijkstra(int s, int t){
    priority_queue<HeapNode>  pq;
    ms(d, INF);
    d[s] = 0;
    ms(done, false);
    pq.push(HeapNode(0, s));

    while(!pq.empty()){
      HeapNode x = pq.top();  pq.pop();
      int u = x.u;
      if(t == x.u)  return d[t];
      if(d[u] >= ans)  return ans;
      if(done[u])  continue;
      done[u] = true;
      for(int i = 0; i < G[u].sz; ++i){
        Edge &e = edges[G[u][i]];
        if(d[e.to] > d[u] + e.dist){
          d[e.to] = d[u] + e.dist;
          pq.push(HeapNode(d[e.to], e.to));
        }
      }
    }
    return d[t];
  }
};

map<P, int> mp;

int cnt;
int ID(const P &p){
  if(mp.count(p))  return mp[p];
  return mp[p] = cnt++;
}
Dijskstra dij;

int main(){
  int T;  cin >> T;
  for(int kase = 1; kase <= T; ++kase){
    scanf("%d", &n);
    mp.cl;  cnt = 0;
    dij.init(n*2+10);
    for(int i = 0; i < n; ++i){
      int x1, y1, x2, y2, w;
      scanf("%d %d %d %d %d", &x1, &y1, &x2, &y2, &w);
      int u = ID(P(x1, y1));
      int v = ID(P(x2, y2));
      dij.addEdge(u, v, w);
      dij.addEdge(v, u, w);
    }
    ans = INF;
    for(int i = 0; i < dij.edges.sz; i += 2){
      int val = dij.edges[i].dist;
      dij.edges[i].dist = dij.edges[i^1].dist = INF;
      ans = min(ans, dij.dijkstra(dij.edges[i].from, dij.edges[i].to) + val);
    }
    printf("Case #%d: %d
", kase, ans == INF ? 0 : ans);
  }
  return 0;
}