poj 1328 - Radar Installation

这道题还是有点意思的，关键在把问题进行转换，说实话使用贪心解决这道题还真不是我想出来的，关键就在想到要求出每个坐标下，雷达所设的最大左和最右位置，然后再进行贪心选择。

思路：求出每个坐标下的雷达的最左和最右区间，对其进行非降序排序，选取第一个区间的的右边CurrentRight作为预设的第一个雷达位置，倘若下一个区间和当前有交集，则雷达可设置在交集内，此时更新CurrentRight；如果没有交集则说明需要新加一个雷达。

/*----------------------------------------------------------------------------------------
* FileName:POJ1328.c
* author:doodlesomething@163.com
* date:10-29-2014
* version:1.0
* description:Radar Installation  POJ1328 soloved by Greedy Algorithm
-----------------------------------------------------------------------------------------*/

#include <stdio.h>
#include <math.h>

#define N 10

typedef struct {
double left;
double right;
}Pos;

//这样设置是为了理解上的方便
typedef struct {
int x;
int y;
}Posi;

/*
* @description:quick sort by the left asc
*/
void QuickSort(Pos P[],int low,int high) {
int first,last;
Pos temp;

if(low >= high) {
return;
}

first = low;
last = high;
temp = P[0];

while(first < last) {
while(first < last && temp.left <= P[last].left)
last--;
P[first] = P[last];

while(first < last && temp.left >= P[first].left)
first++;
P[last] = P[first];
}

P[first] = temp;

QuickSort(P,low,first - 1);
QuickSort(P,first + 1,high);

}

int main() {

int n,dis,i,flag,total;
double CurrentRight;
Pos P[1000];
Posi S[1000];

flag = 1;
total = 1;

printf("please enter n and dis:");
scanf("%d,%d",&n,&dis);
printf("please enter each value:\n");
for(i = 0; i < n; i++) {
scanf("%d%d",&S[i].x,&S[i].y);
if(S[i].y > dis)
flag = 0;
}

if(!flag) {
printf("result:-1\n");
}
else {
for(i = 0; i < n ; i++) {
P[i].left =  S[i].x - sqrt( (double) (dis * dis - S[i].y * S[i].y));
P[i].right = S[i].x + sqrt( (double) (dis * dis - S[i].y * S[i].y));
}

//QuickSort
QuickSort(P,0,n - 1);

CurrentRight = P[0].right;

for( i = 1; i < n; i ++) {
if(P[i].left > CurrentRight) {
total++;
CurrentRight = P[i].right;
}
else {
CurrentRight = P[i].right >  CurrentRight ? CurrentRight : P[i].right;
}
}
//output
printf("%d\n",total);
}

return 0;
}

/*
Sample Input

3,2
1 2
-3 1
2 1

Sample Output
total:2

Sample Input
1,2
0 2

Sample Output
total:1

*/