Travel in Desert UVA

Travel in Desert UVA - 10816

图论·生成树·最短路

http://blog.csdn.net/weizhuwyzc000/article/details/50674893

题目大意：

有n个绿洲， m条道路，每条路上有一个温度，和一个路程长度，从绿洲s到绿洲t，求一条道路的最高温度尽量小， 如果有多条， 选一条总路程最短的。

题解：

按温度跑最小生成树，当s和t连通的时候当前边的温度就是最小温度，然后把所有小于等于这个温度的边都加进来，跑权值的最短路即可。

或者二分答案判断是否连通也行。

Code：

#include<cstdio>  #include<cstring>  #include<algorithm>  #include<iostream>  #include<string>  #include<vector>  #include<stack>  #include<bitset>  #include<cstdlib>  #include<cmath>  #include<set>  #include<list>  #include<deque>  #include<map>  #include<queue>  #define Max(a,b) ((a)>(b)?(a):(b))  #define Min(a,b) ((a)<(b)?(a):(b))  using namespace std;  typedef long long ll;  const double PI = acos(-1.0);  const double eps = 1e-6;  const int mod = 1000000000 + 7;  const double INF = 1000000000.0;  const int maxn = 100 + 10;  int T,n,m,S,p[maxn],path[maxn];  double ans_t,ans_l,d[maxn];  bool done[maxn], vis[maxn*maxn];  struct node {      int a, b;      double r, d;      node(int a=0,int b=0,double r=0,double d=0):a(a),b(b),r(r),d(d) {}      bool operator < (const node& rhs) const {          return r < rhs.r;      }  }a[maxn*maxn];  struct hehe {      int a;      double d;      hehe(int a=0, double d=0):a(a), d(d) {}      bool operator < (const hehe& rhs) const {          return d > rhs.d;      }  };  vector<node> g[maxn];  int _find(int x) { return p[x] == x ? x : p[x] = _find(p[x]); }  void print(int root) {      vector<int> ans;      while(root != S) {          ans.push_back(root);          root = path[root];      }      ans.push_back(S);      int len = ans.size();      for(int i=len-1;i>=0;i--) {          if(i == len-1) printf("%d",ans[i]);          else printf(" %d",ans[i]);      }      printf("\n%.1f %.1f\n",d[T],ans_t);  }  void dij() {      priority_queue<hehe> q;      for(int i=1;i<=n;i++) d[i] = INF, done[i] = false;      d[S] = 0;      q.push(hehe(S,0));      while(!q.empty()) {          hehe u = q.top(); q.pop();          if(done[u.a]) continue;          done[u.a] = true;          int len = g[u.a].size();          for(int i=0;i<len;i++) {              node v = g[u.a][i];              if(d[v.a] > d[u.a] + v.d && fabs(d[v.a]-d[u.a]-v.d) > eps) {                  d[v.a] = d[u.a] + v.d;                  path[v.a] = u.a;                  q.push(hehe(v.a,d[v.a]));              }          }      }      print(T);  }  void solve() {      sort(a, a+m);      for(int i=1;i<=n;i++) p[i] = i, g[i].clear();      ans_t = 0.0;      memset(vis, false, sizeof(vis));      for(int i=0;i<m;i++) {          int x = _find(a[i].a), y = _find(a[i].b);          if(x != y) {              p[x] = y;              vis[i] = true;              g[a[i].a].push_back(node(a[i].b,0,a[i].r,a[i].d));              g[a[i].b].push_back(node(a[i].a,0,a[i].r,a[i].d));              x = _find(S);              y = _find(T);              ans_t = max(ans_t, a[i].r);              if(x == y) break;          }      }      for(int i=0;i<m;i++) {          if(!vis[i] && (ans_t > a[i].r || fabs(a[i].r-ans_t) < eps)) {              g[a[i].a].push_back(node(a[i].b,0,a[i].r,a[i].d));              g[a[i].b].push_back(node(a[i].a,0,a[i].r,a[i].d));          }      }      dij();  }  int main() {      while(~scanf("%d%d",&n,&m)) {          scanf("%d%d",&S,&T);          for(int i=0;i<m;i++) scanf("%d%d%lf%lf",&a[i].a,&a[i].b,&a[i].r,&a[i].d);          solve();      }      return 0;  }