Floyd-warshall algorithm (solves the shortest path between any two points) detailed + deformation of the POJ 2253 Frogger

/* For a long time did not do the problem of graph theory, review. ---------------------------------------------------------Shortest-circuiting (Floyd-warshall algorithm) dynamic programming between any two points: dp[k][i][j] :=   node I can reach the shortest path of the J-node through a node numbered 1,2...K. The use of 1,2...K nodes, can be divided into the following two scenarios to discuss: (1) I to j the shortest path through the node K once Dp[k-1][i][k] + dp[k-1][k][j] (2) I to j the shortest path completely without the node Kdp[k-1] I [j] therefore: Dp[k][i][j] = min (Dp[k-1][i][k] + dp[k-1][k][j], dp[k-1][i][j]) space is optimized: Dp[i][j]  = min (Dp[i][k] + dp[k][j], dp[i][j]) initial value:dp[0][i][j] = 0  (i ==  J) The weight of the dp[0][i][j] :=  edge i->j if there is no edge,dp[i][j] = inf  (0x3f3f3f3f) time complexity between I and J: O (| V|^3) can handle the case that the edge is negative, judge whether there is a negative circle in the graph, just check if there is a dp[i][i] is a negative vertex I can be. ---------------------------------------------------------------------------------poj 2253  Frogger for this problem, ask: To execute a given sequence of jumps, a frog ' s  jump range obviously must be at least as long as the  Longest jump oCcuring in the sequence. the frog distance  (humans also call it minimax distance)  between  two stones therefore is defined asthe minimum necessary jump  range over all possible paths between the two stones.  namely:a path needs to jump many times, but there is a jump in all jumps need to jump the longest distance, this distance can be called the longest distance to jump in this path, find a stone to reach the B stone all the path of the need to jump the longest distance, where the shortest distance is the answer, namely: the Frog Distance The Floyd-warshall algorithm traverses all paths to the i->j, so we can change the DP state to solve this problem: dp[k][i][j]: = node I can reach the J-node path through the numbered 1,2...K node, the shortest maximum distance required to jump. So: dp[k][i][j] = min (Dp[k-1][i][j], Max (Dp[k-1][i][k], dp[k-1][k][j])) Optimize: dp[i][j]: = Min (Dp[i][j], Max (dp[i][k), dp[k ][J])) Initial value: Dp[0][i][j]: = Distance from node i to node J. */

1#include <iostream>2#include <cstdlib>3#include <cstdio>4#include <cstddef>5#include <iterator>6#include <algorithm>7#include <string>8#include <locale>9#include <cmath>Ten#include <vector> One#include <cstring> A#include <map> -#include <utility> -#include <queue> the#include <stack> -#include <Set> - using namespacestd; - Const intINF =0x3f3f3f3f; + Const intMAXN =205; - Const intModprime =3046721; +  A structNode at { -     Doublex, y; - }; -  - intN; - Node NODESET[MAXN]; in DoubleDP[MAXN][MAXN]; - DoubleDIS[MAXN][MAXN]; to  + voidgetdistance () - { the      for(inti =1; I <= N; ++i) *     { $          for(intj =1; J <= N; ++j)Panax Notoginseng         { -             if(I! =j) the             {     +DIS[I][J] = sqrt ((nodeset[i].x-nodeset[j].x) * (nodeset[i].x-nodeset[j].x) + A(NODESET[I].Y-NODESET[J].Y) * (NODESET[I].Y-nodeset[j].y)); theDis[j][i] =Dis[i][j]; +             } -             Else $             { $DIS[I][J] =0.0; -             } -         } the     } - }Wuyi  the voidSolve () - { Wu      for(inti =1; I <= N; ++i) -     { About          for(intj =1; J <= N; ++j) $         { -DP[I][J] =Dis[i][j]; -         } -     } A      for(intK =1; K <= N; ++k) +     { the          for(inti =1; I <= N; ++i) -         { $              for(intj =1; J <= N; ++j) the             { theDP[I][J] =min (dp[i][j], Max (Dp[i][k], dp[k][j])); the             } the         } -     } inprintf"Frog Distance =%.3lf\n\n", dp[1][2]); the  the } About  the intMain () the { the #ifdef HOME +Freopen ("inch","R", stdin); -     //freopen ("Out", "w", stdout); the #endifBayi  the     intnum =1; the      while(~SCANF ("%d", &n) &&N) -     { -          for(inti =1; I <= N; ++i) the         { thescanf"%LF%LF", &nodeset[i].x, &nodeset[i].y); the         } the getdistance (); -printf"Scenario #%d\n", num); the Solve (); the++num; the     }94  the  the #ifdef HOME theCerr <<"Time Elapsed:"<< clock ()/Clocks_per_sec <<"Ms"<<Endl;98 _CrtDumpMemoryLeaks (); About #endif -     return 0;101}

Floyd-warshall algorithm (solves the shortest path between any two points) detailed + deformation of the POJ 2253 Frogger