Prisoner Escape

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Prisoner Escape Course Course outline Week 1: Introduction

Due on 2016-08-12, 23:59 IST

Prisoner Escape (Baltic Olympiad in Informatics, 2009)

Week 1: Analysis of algorithms Week 1 Quiz Week 2: Searching and sorting Week 2 Quiz

A group of war prisoners are trying to escape from a prison. They have thoroughly planned the escape from the prison itself, and after that they hope to find shelter in a nearby village. However, the village (marked as B, see picture below) and the prison (marked as A) are separated by a canyon which is also guarded by soldiers. These soldiers sit in their pickets and rarely walk; the range of view of each soldier is limited to exactly 100 meters. Thus, depending on the locations of soldiers, it may be possible to pass the canyon safely, keeping the distance to the closest soldier strictly larger than 100 meters at any moment.

Week 2 Programming Assignment Week 3: Graphs Week 3 Quiz Week 3 Programming Assignment Prisoner Escape

Week 4: Weighted graphs Week 4 Quiz Week 4 Programming Assignment Week 5: Data Structures: Union-Find and Heaps

You are to write a program which, given the width and the length of the canyon and the coordinates of every soldier in the canyon, and assuming that soldiers do not change their locations, determines whether prisoners can pass the canyon unnoticed.

Solution Hint The soldiers can be modelled as an undirected graph G. Let each soldier be represented by a vertex. Add an edge between two vertices if and only if the range of view of the corresponding soldiers overlaps. Add two additional vertices s and t, representing the northern and southern side of the canyon, respectively. Connect s and t to those vertices representing soldiers who range of view includes the respective side of the canyon. Use depth-firstsearch or breadth-first-search to check whether there is a path between s

https://onl i necour ses.nptel .ac.i n/noc16_cs22/pr ogassi gnm ent?nam e= 112

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Week 5: Divide and Conquer Week 5 Quiz Week 5 Programming Assignment

Design and analysis of algorithms - Course

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Input format The first line contains three integers L, W, and N – the length and the width of the canyon, and the number of soldiers, respectively. Each of the following N lines contains a pair of integers Xi and Yi – the coordinates of ith soldier in the canyon (0 ≤ Xi ≤ L, 0 ≤ Yi ≤ W). The coordinates are given in meters, relative to the canyon: the southwestern corner of the canyon has coordinates (0, 0), and the northeastern corner of the canyon has coordinates (L,W), as seen in the picture above. Note that passing the canyon may start at coordinate (0,ys) for any 0 ≤ ys ≤ W and end at coordinate (L,ye) for any 0 &\le; ye ≤ W. Neither ys nor ye need to be integer.

Output format Output a single integer: 0 if the prisoners can escape, 1 if they cannot.

Test data 1≤W ≤ 50,000 1 ≤ L ≤ 50,000 1 ≤ N ≤ 250

Example Sample input 1 130 340 5 10 50 130 130 70 170 0 180 60 260

Sample output 1 1

Sample input 2 500 250 250 100 400

300 4 0 300 150 150

Sample output 2 0 Sample Test Cases Input 130 340 5 10 50 Test Case 1 https://onlinecourses.nptel.ac.in/noc16_cs22/progassignment?name=112

130 130 70 170

Output

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0 180 60 260 500 300 6 100 0 100 150

Test Case 2

100 300

1

400 0 400 150 400 300 500 300 5 250 0

Test Case 3

250 150 250 300

1

100 150 400 150 500 300 4 250 0

Test Case 4

250 300

0

100 150 400 150 500 300 5 50 25

Test Case 5

450 25 100 275

0

250 120 400 275 500 300 6 250 0 100 300

Test Case 6

200 150

1

300 150 400 300 250 300

Due Date Exceeded. You scored 100.0/100. Your last recorded submission was :

https://onlinecourses.nptel.ac.in/noc16_cs22/progassignment?name=112

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#include #include #include #include #include

using namespace std;

#define sq(x) ((x)*(x)) struct pt:pair { int i; pt(int x=0,int y=0):pair(x,y),i(‐1) { };

}

int main() { int L,W,n,r=100,d=2*r; cin>>L>>W>>n;

set

points;

typedef set::iterator iter; pt P;

list G[n+2]; //Prepare Graph with intelligence {using lgn based set} for(int i=1; i<=n; i++) { cin>>P.first>>P.second; P.i=i; pt Q(P.first‐d,P.second‐d),R(P.first+ d,P.second+d); iter l=points.lower_bound(Q),u=points.upper_bound(R); for(iter j=l; j!=u; ++j) // only those circles which can overl if(sq(P.first‐j‐>first)+sq(P.second‐j‐>second) <= sq(d)) G[i].push_back(j‐>i), G[j‐>i].push_back( i); // if P is touching y=0 if(P.second ‐ r <= 0) G[i].push_back(0), G[0].push_back(i); if(P.second + r >= W) // if P is touching y=W G[i].push_back(n+1), G[n+1].push_back(i);

}

points.insert(P);

//Do DFS/BFS to track int visited[n+2]; for(int i=0;i

deque q; q.push_back(0); // the y=0 line

while (!q.empty()) { int v=q.front(); q.pop_front(); if(visited[v]) continue ; visited[v]=1; if(v==n+1) //the y=L line { cout<<1<<"\n"; //there is an path from y=0 to y=L return 0; } for(list::iterator i=G[v].begin(); i != G[v].end(); ++i) q.push_back(*i);

}

}

cout<<0<<"\n"; return 0;

End


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