POJ 3069 Saruman's Army
2018-01-31 16:58
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Saruman's Army
Description
Saruman the White must lead his army along a straight path from Isengard to Helm’s Deep. To keep track of his forces, Saruman distributes seeing stones, known as palantirs, among the troops. Each palantir has a maximum effective range of R units,
and must be carried by some troop in the army (i.e., palantirs are not allowed to “free float” in mid-air). Help Saruman take control of Middle Earth by determining the minimum number of palantirs needed for Saruman to ensure that each of his minions is within R units
of some palantir.
Input
The input test file will contain multiple cases. Each test case begins with a single line containing an integer R, the maximum effective range of all palantirs (where 0 ≤ R ≤ 1000), and an integer n, the number of troops in Saruman’s
army (where 1 ≤ n ≤ 1000). The next line contains n integers, indicating the positions x1, …, xn of each troop (where 0 ≤ xi ≤ 1000). The end-of-file is marked by a test case with R = n =
−1.
Output
For each test case, print a single integer indicating the minimum number of palantirs needed.
Sample Input
Sample Output
Hint
In the first test case, Saruman may place a palantir at positions 10 and 20. Here, note that a single palantir with range 0 can cover both of the troops at position 20.
In the second test case, Saruman can place palantirs at position 7 (covering troops at 1, 7, and 15), position 20 (covering positions 20 and 30), position 50, and position 70. Here, note that palantirs must be distributed among troops and are not allowed
to “free float.” Thus, Saruman cannot place a palantir at position 60 to cover the troops at positions 50 and 70.
简单贪心
我的代码:
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 12362 | Accepted: 6226 |
Saruman the White must lead his army along a straight path from Isengard to Helm’s Deep. To keep track of his forces, Saruman distributes seeing stones, known as palantirs, among the troops. Each palantir has a maximum effective range of R units,
and must be carried by some troop in the army (i.e., palantirs are not allowed to “free float” in mid-air). Help Saruman take control of Middle Earth by determining the minimum number of palantirs needed for Saruman to ensure that each of his minions is within R units
of some palantir.
Input
The input test file will contain multiple cases. Each test case begins with a single line containing an integer R, the maximum effective range of all palantirs (where 0 ≤ R ≤ 1000), and an integer n, the number of troops in Saruman’s
army (where 1 ≤ n ≤ 1000). The next line contains n integers, indicating the positions x1, …, xn of each troop (where 0 ≤ xi ≤ 1000). The end-of-file is marked by a test case with R = n =
−1.
Output
For each test case, print a single integer indicating the minimum number of palantirs needed.
Sample Input
0 3 10 20 20 10 7 70 30 1 7 15 20 50 -1 -1
Sample Output
2 4
Hint
In the first test case, Saruman may place a palantir at positions 10 and 20. Here, note that a single palantir with range 0 can cover both of the troops at position 20.
In the second test case, Saruman can place palantirs at position 7 (covering troops at 1, 7, and 15), position 20 (covering positions 20 and 30), position 50, and position 70. Here, note that palantirs must be distributed among troops and are not allowed
to “free float.” Thus, Saruman cannot place a palantir at position 60 to cover the troops at positions 50 and 70.
简单贪心
我的代码:
//#include <bits/stdc++.h> #include <stdio.h> #include <stdlib.h> #include <string.h> #include <iostream> #include <algorithm> using namespace std; const int maxn = 1e4; bool used[maxn]; int perm[maxn]; char s[maxn]; int n; int main() { int r,n; //freopen("/Users/vector/Desktop/input.txt","r",stdin); while(~scanf("%d%d",&r,&n)) { int cnt = 0; if(r==-1&&n==-1)break; for(int i = 0; i < n; i++) scanf("%d",&perm[i]); sort(perm,perm+n);//排序 for(int i = 0; i < n;) { int j = i; while(perm[i] + r >= perm[j+1] && j < n) j++;//向前探测 i = j; while(perm[i] + r >= perm[j+1] && j < n) j++;//向后探测看能包括几个点 cnt++; if(j >= n)break; i = j + 1; } printf("%d\n",cnt); } return 0; }《挑战》代码:
Source Code Problem: 3069 User: 10152150116 Memory: 708K Time: 16MS Language: G++ Result: Accepted Source Code #include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <vector> #include <cmath> #include <cstring> #include <string> #define N 1e5+10 /* run this program using the console pauser or add your own getch, system("pause") or input loop */ using namespace std; int main(int argc, char** argv) { int r,n; while(~scanf("%d%d",&r,&n)) { int loc[1010]={0}; if(n==-1&&r==-1)break; for(int i=0;i<n;i++) scanf("%d",&loc[i]); sort(loc,loc+n); int index=0,cnt=0; while(index<n) { int s=loc[index]; while(loc[index]-s<=r) index++; s=loc[index-1]; while(s+r>=loc[index-1]) index++; index--; cnt++; } printf("%d\n",cnt); } return 0; }
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