hdoj 1596 find the safest road

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1596

解题思路:Floyd 算法(邻接矩阵)

  1 /**************************************************************************
  2 user_id: SCNU20102200088
  3 problem_id: hdoj 1596
  4 problem_name: find the safest road
  5 **************************************************************************/
  6 
  7 #include <algorithm>
  8 #include <iostream>
  9 #include <iterator>
 10 #include <iomanip>
 11 #include <sstream>
 12 #include <fstream>
 13 #include <cstring>
 14 #include <cstdlib>
 15 #include <climits>
 16 #include <bitset>
 17 #include <string>
 18 #include <vector>
 19 #include <cstdio>
 20 #include <cctype>
 21 #include <ctime>
 22 #include <cmath>
 23 #include <queue>
 24 #include <stack>
 25 #include <list>
 26 #include <set>
 27 #include <map>
 28 using namespace std;
 29 
 30 //线段树
 31 #define lson l,m,rt<<1
 32 #define rson m+1,r,rt<<1|1
 33 
 34 //手工扩展栈
 35 #pragma comment(linker,"/STACK:102400000,102400000")
 36 
 37 const double EPS=1e-9;
 38 const double PI=acos(-1.0);
 39 const double E=2.7182818284590452353602874713526;  //自然对数底数
 40 const double R=0.5772156649015328606065120900824;  //欧拉常数:(1+1/2+...+1/n)-ln(n)
 41 
 42 const int x4[]={-1,0,1,0};
 43 const int y4[]={0,1,0,-1};
 44 const int x8[]={-1,-1,0,1,1,1,0,-1};
 45 const int y8[]={0,1,1,1,0,-1,-1,-1};
 46 
 47 typedef long long LL;
 48 
 49 typedef int T;
 50 T max(T a,T b){ return a>b? a:b; }
 51 T min(T a,T b){ return a<b? a:b; }
 52 T gcd(T a,T b){ return b==0? a:gcd(b,a%b); }
 53 T lcm(T a,T b){ return a/gcd(a,b)*b; }
 54 
 55 ///////////////////////////////////////////////////////////////////////////
 56 //Add Code:
 57 const int maxn=1005;
 58 int n;
 59 double dist[maxn][maxn],f[maxn][maxn];
 60 
 61 void Floyd(){
 62     int i,j,k;
 63     for(i=1;i<=n;i++){
 64         for(j=1;j<=n;j++) dist[i][j]=f[i][j];
 65     }
 66     for(k=1;k<=n;k++){
 67         for(i=1;i<=n;i++){
 68             for(j=1;j<=n;j++){
 69                 if(dist[i][k]*dist[k][j]>dist[i][j]) dist[i][j]=dist[i][k]*dist[k][j];
 70             }
 71         }
 72     }
 73 }
 74 ///////////////////////////////////////////////////////////////////////////
 75 
 76 int main(){
 77     std::ios::sync_with_stdio(false);
 78     //freopen("in.txt","r",stdin);
 79     //freopen("out.txt","w",stdout);
 80     ///////////////////////////////////////////////////////////////////////
 81     //Add Code:
 82     int q,i,j,start,end;
 83     while(scanf("%d",&n)!=EOF){
 84         for(i=1;i<=n;i++){
 85             for(j=1;j<=n;j++) scanf("%lf",&f[i][j]);
 86         }
 87         Floyd();
 88         scanf("%d",&q);
 89         while(q--){
 90             scanf("%d%d",&start,&end);
 91             if(dist[start][end]==0) printf("What a pity!
");
 92             else printf("%.3lf
",dist[start][end]);
 93         }
 94     }
 95     ///////////////////////////////////////////////////////////////////////
 96     return 0;
 97 }
 98 
 99 /**************************************************************************
100 Testcase:
101 Input:
102 3
103 1 0.5 0.5
104 0.5 1 0.4
105 0.5 0.4 1
106 3
107 1 2
108 2 3
109 1 3
110 Output:
111 0.500
112 0.400
113 0.500
114 **************************************************************************/
原文地址:https://www.cnblogs.com/linqiuwei/p/3359403.html