{POJ}{3897}{Maze Stretching}{二分答案+BFS}

题意：给定迷宫，可以更改高度比，问如何使最短路等于输入数据。

思路：由于是单调的，可以用二分答案，然后BFS验证。这里用优先队列，每次压入也要进行检查（dis大小）防止数据过多，A*也可以。好久不写图论，WA成狗

#include <iostream>
#include <string>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <memory>
#include <cmath>
#include <bitset>
#include <queue>
#include <vector>
#include <stack>
using namespace std;

#define CLR(x,y) memset(x,y,sizeof(x))
#define MIN(m,v) (m)<(v)?(m):(v)
#define MAX(m,v) (m)>(v)?(m):(v)
#define ABS(x) ((x)>0?(x):-(x))
#define rep(i,x,y) for(i=x;i<y;++i)

#define SET_NODE(no,a,b,c) {no.x=a;no.y=b;no.val=c;}

const int MAXN = 200;
const double INF = 1<<30;
const double EPS = 0.000001;

int dir[4][2]={1,0,-1,0,0,1,0,-1};
int n,m;
int tag[MAXN];
char g[MAXN][MAXN];
double dist;
bool visit[MAXN][MAXN];
double dis[MAXN][MAXN];
typedef struct
{
	int x,y;
	double val;
}Node;
Node s,e,node;
bool operator < (const Node& a,const Node& b)
{
    return a.val > b.val;
}
bool check(const int& x, const int& y)
{
	if(x<0||x>=n) return false;
	if(y<0||y>=m) return false;
	if(g[x][y] == ‘#‘) return false;
	if(visit[x][y]) return false;
	return true;
}

double BFS(double len)
{
	int i,j,k;
	Node tmp;

	priority_queue<Node> q;

	q.push(s);

	CLR(visit,0);

	rep(i,0,n)
		rep(j,0,m)
			dis[i][j] = INF;

	while(!q.empty())
	{
		node = q.top();
		q.pop();
		visit[node.x][node.y] = true;

		if(node.x == e.x && node.y == e.y)
			return node.val;

		if(check(node.x+1,node.y)){
			SET_NODE(tmp,node.x+1,node.y,node.val+len);
			if(tmp.val<dis[tmp.x][tmp.y])
			{
				dis[tmp.x][tmp.y] = tmp.val;
				q.push(tmp);
			}
		}
		if(check(node.x-1,node.y)){
			SET_NODE(tmp,node.x-1,node.y,node.val+len);
			if(tmp.val<dis[tmp.x][tmp.y])
			{
				dis[tmp.x][tmp.y] = tmp.val;
				q.push(tmp);
			}
		}
		if(check(node.x,node.y+1)){
			SET_NODE(tmp,node.x,node.y+1,node.val+1);
			if(tmp.val<dis[tmp.x][tmp.y])
			{
				dis[tmp.x][tmp.y] = tmp.val;
				q.push(tmp);
			}
		}
		if(check(node.x,node.y-1)){
			SET_NODE(tmp,node.x,node.y-1,node.val+1);
			if(tmp.val<dis[tmp.x][tmp.y])
			{
				dis[tmp.x][tmp.y] = tmp.val;
				q.push(tmp);
			}
		}
	}
	return 0;
}
void Solve()
{
	int i,j,k,t,tt;

	scanf("%d",&tt);
	rep(t,1,tt+1){
		scanf("%lf%d",&dist,&n);
		rep(i,0,n)
		{
		getchar();
			gets(&g[i][0]);
		}
		m = strlen(g[0]);
		rep(i,0,n)
			rep(j,0,m)
				if(g[i][j]==‘S‘)
				{
					s.x = i;
					s.y = j;
				}else if (g[i][j]==‘E‘)
				{
					e.x = i;
					e.y = j;
				}
		s.val = 0;

		double low = 0;
		double high = 10;
		double mid = (low+high)/2;
		double res = 0;
		while(ABS(low-high)>EPS)
		{
			mid = (low+high)/2;
			res = BFS(mid);
			//printf("[%f %f %f]\n",mid,res,dist);
			if(res < dist+EPS)
				low = mid ;
			else
				high = mid;
		}
		printf("Case #%d: %.3f%%\n",t,mid*100);
	}
}
int main()
{
	Solve();
	return 0;
}