UVa 1382 Distant Galaxy 解题报告（枚举 + 前缀和）

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1382 - Distant Galaxy

Time limit: 3.000 seconds

You are observing a distant galaxy using a telescope above the Astronomy Tower, and you think that a rectangle drawn in that galaxy whose edges are parallel to coordinate axes and contain maximum star systems on its edges has a great deal to do with the mysteries of universe. However you do not have the laptop with you, thus you have written the coordinates of all star systems down on a piece of paper and decide to work out the result later. Can you finish this task?

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Input

There are multiple test cases in the input file. Each test case starts with one integerN , (1 $\le$ N $\le$ 100), the number of star systems on the telescope. N lines follow, each line consists of two integers: theX and Y coordinates of theK-th planet system. The absolute value of any coordinate is no more than10⁹, and you can assume that the planets are arbitrarily distributed in the universe.

N = 0 indicates the end of input file and should not be processed by your program.

Output

For each test case, output the maximum value you have found on a single line in the format as indicated in the sample output.

Sample Input

10 2 3 9 2 7 4 3 4 5 7 1 5 10 4 10 6 11 4 4 6 0

Sample Output

Case 1: 7

解题报告：绝对的好题目。

《训练指南》上说的很清楚了，不说啥了。总之，仔细体会，必有收获。代码如下：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <vector>#include <queue>#include <map>#include <string>using namespace std;#define ff(i, n) for(int i=0;i<(n);i++)#define fff(i, n, m) for(int i=(n);i<=(m);i++)#define dff(i, n, m) for(int i=(n);i>=(m);i--)#define mem(a) memset((a), 0, sizeof(a))typedef long long LL;typedef unsigned long long ULL;void work();int main(){#ifdef ACM    freopen("in.txt", "r", stdin);//    freopen("in.txt", "w", stdout);#endif // ACM    work();}/*****************************************/struct Point{    int x, y;    bool operator<(const Point & cmp) const    {        return x < cmp.x;    }} p[111];int n, m, y[111], l[111], on[111], on2[111];int solve(){    sort(p, p+n);    sort(y, y+n);    m = unique(y, y+n)-y;    if(m <= 2) return n;    int ans = 0;    ff(a, m) fff(b, a+1, m)    {        int miny = y[a], maxy = y[b];        int k = 0;        ff(i, n)        {            if(i==0 || p[i].x != p[i-1].x)            {                k++;                on[k] = on2[k] = 0;                l[k] = l[k-1]+on2[k-1]-on[k-1];            }            if(p[i].y > miny && p[i].y < maxy) on[k]++;            if(p[i].y >= miny && p[i].y <= maxy) on2[k]++;        }        int M = 0;        fff(j, 1, k)        {            ans = max(ans, l[j]+on2[j]+M);            M = max(M, on[j]-l[j]);        }    }    return ans;}void work(){    int cas = 1;    while(~scanf("%d", &n) && n)    {        ff(i, n)        {            scanf("%d%d", &p[i].x, &p[i].y);            y[i] = p[i].y;        }        printf("Case %d: %d\n", cas++, solve());    }}

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