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Escape from Enemy Territory
Description A small group of commandos has infiltrated deep into enemy territory. They have just accomplished their mission and now have to return to their rendezvous point. Of course they don’t want to get caught even if the mission is already over. Therefore they Being well prepared for the mission, they have a detailed map of the area which marks all (known) enemy bases, their current position and the rendezvous point. For simplicity, we view the the map as a rectangular grid with integer coordinates (x, y) Can you help them find the best route? Of course, in case that there are multiple routes that keep the same minimum distance to enemy bases, the commandos want to take a shortest route that does so. Furthermore, they don’t want to take a route off their Input On the first line one positive number: the number of testcases, at most 100. After that per testcase:
All pairs of coordinates are on the map and different from each other. Output Per testcase:
Sample Input 2 1 2 2 0 0 1 1 0 1 2 5 6 0 0 4 0 2 1 2 3 Sample Output 1 2 2 14 Source |
题意：
给你一张X*Y矩形地图。上面有些点上有敌营。给你起点和终点要你找出一条最优路径。满足最优路径上的点离敌营的最近最短距离是所有路径最短的。若有多条找路径最短的一条。
思路：
感觉有的dfs的迭代加深。通过二分来确定路径离敌营最短距离。然后bfs来验证。
详细见代码：
#include<algorithm>
#include<iostream>
#include<string.h>
#include<sstream>
#include<stdio.h>
#include<math.h>
#include<vector>
#include<string>
#include<queue>
#include<set>
#include<map>
//#pragma comment(linker,"/STACK:1024000000,1024000000")
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=1e-8;
const double PI=acos(-1.0);
const int maxn=100010;
typedef __int64 ll;
int n,X,Y,sx,sy,ex,ey,mdis,head,tail,le,ri,ansd,anss;
int vis[1010][1010],dis[1010][1010];
int dx[4]={-1,0,1,0};
int dy[4]={0,1,0,-1};
struct yb
{
int x,y,sp;
} q[1010*1010];//开始队列开小了。wa了
void prebfs()//预处理出每个地方离敌营的最短距离
{
int i,x,y,nx,ny;
head=0;
tail=n;
while(head<tail)
{
x=q[head].x;
y=q[head].y;
for(i=0;i<4;i++)
{
nx=x+dx[i];
ny=y+dy[i];
if(nx>=0&&nx<X&&ny>=0&&ny<Y&&!vis[nx][ny])
{
vis[nx][ny]=1;
dis[nx][ny]=q[head].sp+1;
ri=max(ri,dis[nx][ny]);//离敌营最大距离
q[tail].x=nx;
q[tail].y=ny;
q[tail++].sp=q[head].sp+1;
}
}
head++;
}
}
bool bfs()
{
int i,x,y,nx,ny;
if(dis[sx][sy]<mdis)
return false;
memset(vis,0,sizeof vis);
head=tail=0;
q[tail].x=sx;
q[tail].y=sy;
q[tail++].sp=0;
while(head<tail)
{
x=q[head].x;
y=q[head].y;
if(x==ex&&y==ey)//由于是bfs所以路径一定是最短的。
{
ansd=mdis;
anss=q[head].sp;
return true;
}
for(i=0;i<4;i++)
{
nx=x+dx[i];
ny=y+dy[i];
if(nx>=0&&nx<X&&ny>=0&&ny<Y&&!vis[nx][ny]&&dis[nx][ny]>=mdis)
{
vis[nx][ny]=1;
q[tail].x=nx;
q[tail].y=ny;
q[tail++].sp=q[head].sp+1;
}
}
head++;
}
return false;
}
int main()
{
int cas,i,j,x,y;
scanf("%d",&cas);
while(cas--)
{
scanf("%d%d%d",&n,&X,&Y);
scanf("%d%d%d%d",&sx,&sy,&ex,&ey);
memset(vis,0,sizeof vis);
for(i=0;i<n;i++)
{
scanf("%d%d",&x,&y);
q[i].x=x;
q[i].y=y;
q[i].sp=dis[x][y]=0;
vis[x][y]=1;
}
ri=0;
prebfs();
le=0;
while(le<=ri)//二分最优路径离敌营的最短距离
{
mdis=(le+ri)>>1;
if(bfs())
le=mdis+1;
else
ri=mdis-1;
}
printf("%d %d\n",ansd,anss);
}
return 0;
}