Sicily 1234. Playground

最新推荐文章于 2016-04-24 22:58:00 发布

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最新推荐文章于 2016-04-24 22:58:00 发布

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探讨如何利用半圆形钢丝构建封闭图形的问题。通过比较各钢丝长度总和与最长钢丝长度来判断是否能成功构建。适用于算法竞赛等场景。

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1234. Playground

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

George has K <= 20 steel wires shaped in the form of half-circles, with radii a1, a2, . . . , aK. They can be soldered (connected) at the ends, in any angle. Is it possible for George to make a closed shape out of these wires? He does not have to use all the wires.
The wires can be combined at any angle, but may not intersect. Beware of floating point errors.

Input

Each data set consists of a number 0 < K <= 20 on a line by itself, followed by a line of K space-separated numbers ai. Each number is in the range 0 < ai < 10⁷, and has at most 3 digits after the decimal point.
The input will be terminated by a zero on a line by itself.

Output

For each test case, there should be one word on a line by itself; "YES" if it is possible to make a simple connected figure out of the given arcs, and "NO" if it isn’t.

Sample Input

1
4.000
2
1.000 1.000
3
1.455 2.958 4.424
7
1.230 2.577 3.411 2.968 5.301 4.398 6.777
0

Sample Output

NO
YES
NO
YES

#include <stdio.h>
#include <algorithm>
using namespace std;

//其实可以把圆看成一条线段，这条线段的长度等于圆的半径的两倍
//每次都从中找出最长的一条边，并把其他边的和算出来，假如和比最长边小，说明这条最长边是肯定不行了，扔掉，然后重复

int main() {
    bool is_ok;
    int n, i, j;
    double sum, max, a[21];
    while (scanf("%d", &n) && n) {
        for (i = 0, max = 0, sum = 0, is_ok = false; i < n; i++) {
            scanf("%lf", &a[i]);
        }
        sort(a, a + n);
        for (i = n - 1; i >= 0; i--) {
            max = a[i];
            for (j = 0, sum = 0; j < i; j++) {
                sum += a[j];
            }
            if (sum >= max) {
                is_ok = true;
                break;
            }
        }
        if (is_ok) {
            printf("YES\n");
        } else {
            printf("NO\n");
        }
    }
    return 0;
}