POJ 2263 POJ 1556

2263 the topic meaning is to seek the beginning city to the end city the way of the maximum load capacity, can use Dijkstra or Floyd to solve, I use the Floyd feel more intuitive because only need to change the recursive to W[i][j] = Max (w[i][j],Min(w[i][k],w[k][j])); can get the answer, and Floyd writing is relatively simple but the complexity of N 3 times, but the problem is still able to AC

Source Codeproblem:2263user:lyz963254memory:760k time:32mslanguage:g++result:accepted Source Code #include<iostream>#include<cstdio>#include<cstring>using namespacestd; intMax (intXinty) {returnX>y?x:y; }    intMin (intXinty) {returnX<y?x:y; }    Const intM = -; intKase =0; intN,r; intnum; intW[m][m]; Charcity[m][ *]; Charstart[ *],dest[ *]; intIndexChar*s) {inti;  for(i=0; i<num;i++){            if(!STRCMP (City[i],s))returni;        } strcpy (City[i],s); Num++; returni; }    intRead () {intI,j,k,limit; scanf ("%d%d",&n,&R); if(n==0)return 0;  for(i=0; i<n;i++){             for(j=0; j<n;j++) W[i][j]=0; W[i][i]=9999999; } num=0;  for(k=0; k<r;k++) {scanf ("%s%s%d",start,dest,&limit); I=index (START); J=index (dest); W[I][J]= W[j][i] =limit; } scanf ("%s%s", start,dest); return 1; }    voidsolve () {inti,j,k;  for(k=0; k<n;k++){             for(i=0; i<n;i++){                 for(j=0; j<n;j++) {W[i][j]=Max (W[i][j],min (w[i][k],w[k][j)); }}} I=index (START); J=index (dest); printf ("Scenario #%d\n%d tons\n\n",++Kase,w[i][j]); }    intMain () { while(Read ()) {solve (); }        return 0; }

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1556 almost built to build a drunk, the algorithm directly apply Bellman-ford but the process of building the map is too disgusting, you need to judge whether the wall is blocked a, a, a, a vertex, blocking is not connected edge or even edge ...

#include <cstdio>#include<iostream>#include<cstring>#include<cmath>using namespacestd;#defineINF 20000000000#defineM 150structpoint{Doublex; Doubley;} P[M];structedge{intu; intv;} e[10050];intN//the number of roomsintPSize;//Number of pointsDoublewx[ -];//The x-coordinate of the wallDoublepy[ -][4];//coordinates of yDoubleMAP[M][M];//adjacency MatrixintEsize;//the number of edgesinti,j;DoubleDis (Point A,point b) {returnsqrt ((a.x-b.x) * (a.x-b.x) + (A.Y-B.Y) * (a.y-b.y));}DoubleCross (DoubleX1,DoubleY1,DoubleX2,DoubleY2,DoubleX3,DoubleY3)//determine if the point is above or below the line{    return(x2-x1) * (y3-y1)-(x3-x1) * (y2-y1);}BOOLOk (Point A,point B)//determine if two points can be connected at the time of constructing the net.{    if(a.x>=b.x)return false; BOOLFlag =true; inti =0;  while(wx[i]<=a.x&&i<N) {i++; }     while(wx[i]<b.x&&i<N) {        if(Cross (A.x,a.y,b.x,b.y,wx[i),0) *cross (a.x,a.y,b.x,b.y,wx[i],py[i][0]) <0||Cross (a.x,a.y,b.x,b.y,wx[i],py[i][1]) *cross (a.x,a.y,b.x,b.y,wx[i],py[i][2]) <0||Cross (a.x,a.y,b.x,b.y,wx[i],py[i][3]) *cross (A.x,a.y,b.x,b.y,wx[i],Ten) <0) {flag=false;  Break; } I++; }    returnFlag;}DoubleBellmanintBegintend) {    DoubleDist[m]; inti,j;  for(i=0; i<m;i++) {Dist[i]=inf; } Dist[beg]=0; BOOLex =true;  for(i=0; i<psize&&ex;i++) {ex=false;  for(j=0; j<esize;j++){                if(dist[e[j].u]<inf&& (Dist[e[j].v]> (dist[e[j].u]+MAP[E[J].U][E[J].V]))) {DIST[E[J].V]= dist[e[j].u]+MAP[E[J].U][E[J].V]; Ex=true; }        }    }    returndist[end];}voidsolve () {p[0].x =0; p[0].Y =5; PSize=1;  for(i=0; i<n;i++) {scanf ("%LF",&Wx[i]);  for(j=0;j<4; j + +) {p[psize].x=Wx[i]; scanf ("%LF",&p[psize].y); PY[I][J]=p[psize].y; PSize++; }} p[psize].x=Ten; P[psize].y=5; PSize++;  for(i=0; i<psize;i++){             for(j=0; j<psize;j++) {Map[i][j]=inf; }} esize=0;  for(i=0; i<psize;i++){         for(j=i+1; j<psize;j++){            if(Ok (P[i],p[j])) {Map[i][j]=Dis (P[i],p[j]); E[ESIZE].U=i; E[ESIZE].V=J; Esize++; }}} printf ("%.2lf\n", Bellman (0, psize-1));}intMain () { while(SCANF ("%d", &n)! =EOF) {        if(n==-1) Break;    Solve (); }    return 0;}

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POJ 2263 POJ 1556