Poj 1113 Wall (convex hull), poj1113wall convex hull

Poj 1113 Wall (convex hull), poj1113wall convex hull

Link: poj 1113

Given n vertices of a polygon Castle, a wall is built around the outside of the castle to enclose all vertices,

And the distance between the wall and all vertices is at least L. Find the minimum length of the wall.

Idea: minimum length = the total side length of the convex hull formed by the castle vertex + the circle perimeter with a radius of L

Use the Graham algorithm to obtain the convex hull, and then enumerate its vertices to find the edge length between the two. Remember to add the edge length of the first vertex and the last vertex.

The rounded integer result can be output using double storage and %. 0lf output.

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;struct stu{    int x,y;}p[1010],s[1010];int m,top;int chaji(struct stu p1,struct stu p2,struct stu p3){    return (p1.x-p2.x)*(p3.y-p2.y)-(p3.x-p2.x)*(p1.y-p2.y);}double dis(struct stu p1,struct stu p2){    return sqrt((p1.x-p2.x)*(p1.x-p2.x)*1.0+(p1.y-p2.y)*(p1.y-p2.y));}int cmp(struct stu p1,struct stu p2){    int k;    k=chaji(p1,p[0],p2);    if(k>0||(k==0&&dis(p1,p[0])<dis(p2,p[0])))        return 1;    return 0;}void graham(){    struct stu t;    int k=0,i;    for(i=1;i<m;i++)        if(p[i].y<p[k].y||(p[i].y==p[k].y&&p[i].x<p[k].x))            k=i;    t=p[k];    p[k]=p[0];    p[0]=t;    sort(p+1,p+m,cmp);    s[0]=p[0];    s[1]=p[1];    top=1;    for(i=2;i<m;i++){        while(top>=1&&chaji(s[top-1],s[top],p[i])>=0)            top--;        top++;        s[top]=p[i];    }}int main(){    int n,i;    double sum;    while(scanf("%d%d",&m,&n)!=EOF){        for(i=0;i<m;i++)            scanf("%d%d",&p[i].x,&p[i].y);        graham();        sum=dis(s[0],s[top])+2*3.1415926*n;        for(i=1;i<=top;i++)            sum+=dis(s[i-1],s[i]);        printf("%.0lf\n",sum);    }    return 0;}

Poj1113 data

The data of POJ does not seem to be exposed ..
 
ACM solution report

Category of a poj question
Mainstream algorithms:
1. Search // trace back
2. DP (Dynamic Planning)
3. Greedy
4. Graph Theory // Dijkstra, Minimum Spanning Tree, and network stream
5. Number Theory // solves the modulus linear equation
6. Calculate the area and perimeter of the joint of the geometric/Convex Shell with the same placement of the rectangle
7. Composite math // Polya Theorem
8. Simulation
9. Data Structure // check the collection and heap
10. Game Theory
1. Sorting
1423,169 4, 1723,172 7, 1763,178 8, 1828,183 8, 1840,220 1, 2376,
2377,238 0, 1318,187 1971,197, 1990,200 4, 2002,209 1, 2379 2,
1002 (character processing is required, and sorting can be done in a fast way) 1007 (stable sorting) 2159 (difficult to understand)
2231 2371 (simple sorting) 2388 (sequence statistics algorithm) 2418 (Binary sorting tree)
2. Search, backtracking, and traversal
2329
Simple: 1128,116 6, 1176,123 1, 1256,127 0, 1321,154 3, 1606,166 4,
1731,174 2, 1745,184, 1950,203, 2157,218 8, 2183,238 2, 2386,242 1, 6
Not easy: 1024,105 4, 1117,116 7, 1708,174 6, 1775,187 8, 1903,196 6, 2046,
2197,234 9
Recommended: 1011,119 0, 1191,141 6, 1579,163 2, 1639,165 9, 1680,168 3, 1691,
1709,171 4, 1753,1771, 1826,185 5, 1856,189 0, 1924,193 5, 1948,197 9, 1980,217 2331,233 1979, 1980 9, (similar to the maze), (higher requirements for pruning)
3. Calendar
1008 2080 (Be careful with such questions)
4. Enumeration
1387,141, 2245,232, 2363,238 1, 1650 6, (higher pruning requirements), (decimal precision problem)
5. Typical data structure Algorithms
Easy: 1182,165 6, 2021,202 3, 2051,215 3, 2227,223 6, 2247,235 2,
2395
Not easy: 1145,117 7, 1195,122 7, 1661,183 4
Recommended: 1330,133 8, 1451,147 0, 1634,168 9, 1693,170 3, 1724,198 8, 2004,
2010,211 9, 2274
1125 (fresh algorithm), 2421 (Minimum Spanning Tree of the graph)
6. Dynamic Planning
1037 A decorative fence,
1050 To the Max,
1088 skiing,
1125 Stockbroker Grapevine,
1141 Brackets Sequence,
1159 Palindrom ...... remaining full text>