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题意：在w*h的图上有n个点，要求找出一个正方形面积最大，且没有点落在该正方形内部。
析：枚举所有的y坐标，去查找最大矩形，不断更新。
代码如下：
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
using namespace std ;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f3f;
const double eps = 1e-8;
const int maxn = 1e4 + 5;
const int dr[] = {0, 0, -1, 1};
const int dc[] = {-1, 1, 0, 0};
int n, m;
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
struct node{
int x, y;
bool operator < (const node &p) const{
return x < p.x || (x == p.x && y < p.y);
}
};
int d[maxn];
node a[maxn];
int t, x;
void solve(){
int ans = 0, ansx, ansy;
for(int i = 0; i < x; ++i){
for(int j = i+1; j < x; ++j){
int maxy = d[j], miny = d[i];
int h = maxy - miny, w = 0, tmp = 0;
for(int k = 0; k < t; ++k){
if(a[k].y <= miny || a[k].y >= maxy) continue;
w = a[k].x - tmp;
if(ans < min(w, h)){
ans = min(w, h);
ansx = tmp; ansy = miny;
}
tmp = a[k].x;
}
w = m - tmp;
if(ans < min(w, h)){
ans = min(w, h);
ansx = tmp; ansy = miny;
}
}
}
printf("%d %d %d\n", ansx, ansy, ans);
}
int main(){
int T; cin >> T;
while(T--){
scanf("%d %d %d", &t, &m, &n);
for(int i = 0; i < t; ++i){ scanf("%d %d", &a[i].x, &a[i].y); d[i+1] = a[i].y; }
d[0] = 0; d[t+1] = n;
sort(d, d+t+2);
sort(a, a+t);
x = unique(d, d+t+2) - d;
solve();
if(T) printf("\n");
}
return 0;
}
UVa 1312 Cricket Field （枚举+离散化）
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原文地址：http://www.cnblogs.com/dwtfukgv/p/5732231.html
