Frogger POJ 2253

http://poj.org/problem?id=2253

Test instructions: A frog a wants to jump to the other Frog B, because frog A's jumping range is not so big, so it needs to use some of its surrounding stones to reach the Frog B. Ask you the maximum number of hops in a road that is smaller than the maximum value in other roads. (that is, to ask for at least how much to jump)

#include <stdio.h>#include<string.h>#include<math.h>#include<algorithm>#include<queue>using namespaceStd;typedefLong LongLL;#defineOO 0x3f3f3f3f#defineMAXN 500DoubleMAPS[MAXN][MAXN], DIST[MAXN];intV[MAXN];intN;structnode{intx, y;} A[MAXN];DoubleFintIintj) {    return(Pow (sqrt (Double) a[i].x-a[j].x,2) +pow (Double) A[i].y-a[j].y,2)));}DoubleDij () {memset (V,0,sizeof(v));  for(intI=1; i<=n; i++) Dist[i]= maps[1][i]; dist[1] =0; v[1] =1;  for(intI=1; i<n; i++)    {        intindex; Doublemins =10000000;  for(intj=1; j<=n; J + +)        {            if(!v[j] && Dist[j] <mins) {Index=J; Mins=Dist[j]; }} V[index]=1;  for(intj=1; j<=n; J + +)        {            if(!V[j]) {                DoubleMaxs =max (mins, maps[index][j]); if(DIST[J]&GT;MAXS) Dist[j] =Maxs; }        }    }    returndist[2];}intMain () {intCNT =1;  while(SCANF ("%d", &N), N) { for(intI=1; i<=n; i++) scanf ("%d%d", &a[i].x, &a[i].y);  for(intI=1; i<=n; i++)        {             for(intj=1; j<i; J + +) {Maps[i][j]= Maps[j][i] =F (i, j); } Maps[i][i]=0; }       DoubleAns =Dij (); printf ("Scenario #%d\n", cnt++); printf ("Frog Distance =%.3f\n\n", ans); }    return 0;}

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#include <algorithm>#include<queue>#include<stdio.h>#include<string.h>#include<vector>#include<math.h>using namespacestd;Const intMAXN =205;Const intOO =0xFFFFFFF;structpoint{Doublex, y;} P[MAXN];structnode{inty; DoubleLen; Node (intYDoublelen): Y (y), Len (len) {}};vector<node>G[MAXN];DoubleV[MAXN];//find the distance between two pointsDoubleLen (Point A, point B) {returnsqrt ((a.x-b.x) * (a.x-b.x) + (A.Y-B.Y) * (a.y-b.y));}voidSPFA () {Queue<int>Q; Q.push (1);  while(Q.size ()) {ints =Q.front ();        Q.pop (); intLen =g[s].size ();  for(intI=0; i<len; i++) {node Q=G[s][i]; if(V[s] < V[Q.Y] && Q.len <V[q.y]) {V[Q.Y]= Max (V[s], q.len);//to pick the biggest side of a roadQ.push (Q.Y); }        }    }}intMain () {intN, t=1;  while(SCANF ("%d", &N), N) {intI, J;  for(i=1; i<=n; i++) scanf ("%LF%LF", &p[i].x, &p[i].y);  for(i=1; i<=n; i++)             for(j=1; j<=n; J + +)            {                if(i = =j)Continue; DoubleLen =Len (P[i], p[j]);            G[i].push_back (Node (j, Len)); }         for(i=1; i<=n; i++) V[i]=Oo; v[1] =0;        SPFA (); if(t! =1) printf ("\ n"); printf ("Scenario #%d\n", t++); printf ("Frog Distance =%.3f\n", v[2]);  for(i=1; i<=n; i++) g[i].clear (); }    return 0;

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Frogger POJ 2253