C++博客-syhd142-文章分类-排序

UVA 972 Horizon Line

Fucker — Wed, 29 Sep 2010 08:52:00 GMT

题意：给你两个函数水平范围是[0,10]，定义第三个函数是该区间上两个函数的最大值。水平线就是第三个函数的最小值。

解法：排序模拟就好。

#include <stdio.h>
#include <stdlib.h>
#include <algorithm>

#define N 1005

inline double max(double a, double b)
{
    return a > b ? a : b;
}

struct segement
{
    double height;
    double start, end;
}a[N], b[N];

double step[N];

int main()
{
    int n1, n2, n;
    double sum;
    while(~scanf("%d", &n1))
    {
        n = 1;
        sum = 0;
        for(int i = 0; i < n1; i++)
        {
            scanf("%lf %lf", &a[i].height, &a[i].end);
            a[i].start = sum;
            sum += a[i].end;
            a[i].end = sum;
            step[n++] = sum;
        }
        scanf("%d", &n2);
        sum = 0;
        for(int i = 0; i < n2; i++)
        {
            scanf("%lf %lf", &b[i].height, &b[i].end);
            b[i].start = sum;
            sum += b[i].end;
            b[i].end = sum;
            step[n++] = sum;
        }
        std::sort(step, step + n);
        double ans = 1e9, height1, height2;
        for(int i = 1; i < n; i++)
        {
            for(int j = 0; j < n1; j++)
            if(a[j].start <= step[i - 1] && a[j].end >= step[i])
            {
                height1 = a[j].height;
                break;
            }
            for(int j = 0; j < n2; j++)
            if(b[j].start <= step[i - 1] && b[j].end >= step[i])
            {
                height2 = b[j].height;
                break;
            }
            if(max(height1, height2) < ans) ans = max(height1, height2);
        }
        printf("%.3lf\n", ans);
    }
    return 0;
}

Fucker 2010-09-29 16:52 发表评论

UVA 688 Mobile Phone Coverage

Fucker — Wed, 29 Sep 2010 07:32:00 GMT

经典问题，矩形面积并。

解法：一、矩形分割，每个矩形的两个横坐标和两个纵坐标排序，这样得到2n*2n个区间，对这些区间依次判断是否包含在n个矩形中间即可。

二、扫描线。具体还没实现过。

详见：http://www.algorithmist.com/index.php/UVa_688

#include <stdio.h>
#include <algorithm>

#define N 105

struct rectangle
{
    double x, y;
    double r;
}rec[N];

double x[2 * N], y[2 * N];

int main()
{
    int n, cas = 1;
    while(scanf("%d", &n), n)
    {
        double ans = 0;
        for(int i = 0; i < n; i++)
        {
            scanf("%lf %lf %lf", &rec[i].x, &rec[i].y, &rec[i].r);
            x[i] = rec[i].x - rec[i].r, x[i + n] = rec[i].x + rec[i].r;
            y[i] = rec[i].y - rec[i].r, y[i + n] = rec[i].y + rec[i].r;
        }
        std::sort(x, x + 2 * n);
        std::sort(y, y + 2 * n);
        for(int i = 0; i < 2 * n - 1; i++)
        {
            for(int j = 0; j < 2 * n - 1; j++)
            {
                for(int k = 0; k < n; k++)
                {
                    if(rec[k].x - rec[k].r <= x[i] && rec[k].x + rec[k].r >= x[i + 1])
                    if(rec[k].y - rec[k].r <= y[j] && rec[k].y + rec[k].r >= y[j + 1])
                    {
                        ans += (x[i + 1] - x[i]) * (y[j + 1] - y[j]);
                        break;
                    }
                }
            }
        }
        printf("%d %.2lf\n", cas++, ans);
    }
    return 0;
}

Fucker 2010-09-29 15:32 发表评论

UVA 120 Stacks of Flapjacks

Fucker — Sat, 18 Sep 2010 04:07:00 GMT

依次找最大的烙饼，并翻滚放入适当位置，每个烙饼最多2次即可放好。直接模拟就好了。

思想有了，用代码实现还要想好久。。

#include <stdio.h>
#include <string.h>
#include <algorithm>

int a[35], b[35], c[35];

int main()
{
//    freopen("in", "r", stdin);
    char data[1000];
    int n;
    while(gets(data))
    {
        puts(data);
        char *p = strtok(data, " ");
        n = 0;
        while(p)
        {
            sscanf(p, "%d", &a[++n]);
            b[n] = a[n];
            p = strtok(NULL, " ");
        }
        std::sort(b + 1, b + n + 1);
        for(int i = n; i; i--)
        {
            for(int j = 1; j <= i; j++)
            {
                if(b[i] == a[j] && i != j)
                {
                    if(j == 1) printf("%d ", n - i + 1);
                    else printf("%d %d ", n - j + 1, n - i + 1);
                    int v = 0;
                    for(int k = i; k > j; k--) c[++v] = a[k];
                    for(int k = 1; k <= j; k++) c[++v] = a[k];
                    for(int k = i + 1; k <= n; k++) c[++v] = a[k];
                    for(int k = 1; k <= v; k++) a[k] = c[k];
                    break;
                }
            }
        }
        printf("0\n");
    }
    return 0;
}

Fucker 2010-09-18 12:07 发表评论

UVA 10041 Vito's Family

Fucker — Tue, 29 Jun 2010 11:24:00 GMT

能在水一点么。。。

#include <stdio.h>
#include <algorithm>

#define N 505
int a[N];

int main()
{
    int t, n, mid, sum, tmp;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);
        for(int i = 0; i < n; i++)
            scanf("%d", &a[i]);
        std::sort(a, a + n);
        mid = n >> 1;
        tmp = a[mid];
        sum = 0;
        for(int i = 0; i < n; i++)
            sum += abs(a[i] - tmp);
        printf("%d\n", sum);
    }
    return 0;
}

Fucker 2010-06-29 19:24 发表评论