POJ 2263 POJ 1556

2263 题目意思是求起点城市到终点城市的途径中的最大载重量，可以用Dijkstra或者floyd 来解决，我是用的floyd 感觉更直观 因为只需要将递推式改成w[i][j] = Max(w[i][j],Min(w[i][k],w[k][j]));便可得到答案，而且floyd写法比较简单但是复杂度为n的3次方，不过此题还是能AC

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>

using namespace std;

#define inf 20000000000
#define M 150

struct Point
{
double x;
double y;
}p[M];

struct edge
{
int u;
int v;
}e[10050];

int n;//房间的数目
int pSize; //点的数目
double wX[20];//墙的X坐标
double pY[20][4];//Y的坐标
double map[M][M];//邻接矩阵
int eSize; //边的数目
int i,j;

double Dis(Point a,Point b)
{
return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
}

double Cross(double x1,double y1,double x2,double y2,double x3,double y3) //判断点是否在直线的上方还是下方
{
return (x2-x1)*(y3-y1) - (x3-x1)*(y2-y1);
}
bool Ok(Point a,Point b) //判断在构造有向网的时候两点能否连边
{
if(a.x>=b.x) return false;
bool flag = true;
int i = 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],10)<0){
flag = false;
break;
}
i++;
}
return flag;
}
double bellman(int beg,int end)
{
double dist[M];
int i,j;
for(i=0;i<M;i++){
dist[i] = inf;
}
dist[beg] = 0;
bool ex = 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;
}
}
}
return dist[end];
}

void solve()
{
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=10;
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));

}

int main ()
{
while(scanf("%d",&n)!=EOF){
if(n==-1) break;
solve();
}
return 0;
}

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