SRM465

250pt:

给定50个整数点，范围-500-500之间。然后在这些点上选2个点作为中心，画边长为整数的正方形，并且正方形不能重叠（可以不平行），而且而且边长不同为不同方案。求有多少种方案。。

思路：枚举两个点，计算两点的方案数。

#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
using namespace std;
 
#define PB push_back
#define MP make_pair
 
#define REP(i,n) for(i=0;i<(n);++i)
#define FOR(i,l,h) for(i=(l);i<=(h);++i)
#define FORD(i,h,l) for(i=(h);i>=(l);--i)
#define eps 1e-8
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<double> VD;
typedef long long LL;
typedef pair<int,int> PII;
 
class TurretPlacement
{
        public:
        double dist(int x1, int y1, int x2, int y2){
               double x = x1 - x2;
               double y = y1 - y2;
               return  * sqrt(x * x + y * y);
        }
        long long count(vector <int> x, vector <int> y)
        {
              long long ans = ;
              int n = x.size();
              long long d;
              for (int i = ; i < n; ++i)
                 for (int j = i + ; j < n; ++j){
                     d = floor(dist(x[i], y[i], x[j], y[j]) + eps);
                     ans += (d - ) * d / ;
                 }
              return ans;
        }
};

500pt:

给定最多50个炮台， 50个基地，50个供电厂的位置，并且每个炮台攻击范围为L，而对于每个基地，摧毁它或者摧毁他的供电设施会使他无法工作，求使这些基地都无法工作最少需要的能量为多少（攻击一次能量消耗为距离的平方和）

思路：对于每个基地，摧毁它显然要选择能量最少的炮台，供电厂也同样。那么其实就是一个最小割模型。

对于S，对每个基地i建一条流量为摧毁它能量大小的边

对于T，建一条供电厂j到T，流量为摧毁它能量大小的边

对于基地i与供电厂j，如果供电厂j给基地供电，建i->j,流量为Inf的边，

最后跑一边最大流即可

 // BEGIN CUT HERE
 /*
 
 */
 // END CUT HERE
 #line 7 "GreenWarfare.cpp"
 #include <cstdlib>
 #include <cctype>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <string>
 #include <iostream>
 #include <sstream>
 #include <map>
 #include <set>
 #include <queue>
 #include <stack>
 #include <fstream>
 #include <numeric>
 #include <iomanip>
 #include <bitset>
 #include <list>
 #include <stdexcept>
 #include <functional>
 #include <utility>
 #include <ctime>
 using namespace std;
 
 #define PB push_back
 #define MP make_pair
 #define M0(a) memset(a, 0, sizeof(a))
 #define REP(i,n) for(i=0;i<(n);++i)
 #define FOR(i,l,h) for(i=(l);i<=(h);++i)
 #define FORD(i,h,l) for(i=(h);i>=(l);--i)
 #define maxn 1200
 #define Inf 0x3fffffff
 #define maxm 210000
 typedef vector<int> VI;
 typedef vector<string> VS;
 typedef vector<double> VD;
 typedef long long LL;
 typedef pair<int,int> PII;
 struct oo{
       int y, next, f;
 };
 struct MaxFlow{
        int n, S, T, tot;
        int son[maxn], dist[maxn], gap[maxn];
        oo e[maxm];
 
        int sap(int x, int aug){
            if (x == T) return aug;
            int mind = n, sum = ;
            int y, f;
            for (int p = son[x]; p != -; p = e[p].next){
                   y = e[p].y;
                   if (dist[y] +  == dist[x] && e[p].f){
                        f = sap(y, min(e[p].f, aug - sum));
                        e[p].f -= f;
                        e[p^].f += f;
                        sum += f;
                        if (sum == aug || dist[S] >= n) return sum;
                   }
                   if (e[p].f) mind = min(mind, dist[y]);
            }
            if (!sum){
                if (!(--gap[dist[x]])) dist[S] = n;
                ++gap[dist[x] = mind + ];
            }
            return sum;
        }
 
        void add_edge(int x, int y, int f){
             e[tot].y = y; e[tot].f = f;
             e[tot].next = son[x]; son[x] = tot++;
             e[tot].y = x; e[tot].f = ;
             e[tot].next = son[y]; son[y] = tot++;
        }
 
        void init(int S, int T, int n){
             memset(son, -, sizeof(son));
             tot = ;
             this->S = S, this->T = T, this->n = n;
        }
        int maxflow(){
             M0(gap);
             M0(dist);
             gap[] = n;
             int ans = ;
             while (dist[S] < n) ans += sap(S, Inf);
             return ans;
        }
 } F;
 
 class GreenWarfare
 {
         public:
         int n, m, k;
         int dist(int x1, int y1, int x2, int y2){
              int x = x1 - x2;
              int y = y1 - y2;
              return x * x + y * y;
         }
         int minimumEnergyCost(vector <int> cX, vector <int> cY, vector <int> bX, vector <int> bY, vector <int> pX, vector <int> pY, int SPL)
         {
                 n = cX.size(), m = bY.size(), k = pX.size();
                 F.init(, m + k + , m + k + );
                 //printf("%d %d\n", F.T, F.n);
                 for (int i = ; i < m; ++i)
                    for (int j = ; j < k; ++j)
                        if (dist(bX[i], bY[i], pX[j], pY[j]) <= SPL * SPL) F.add_edge(i + , m + j + , Inf);
                 for (int i = ; i < m; ++i){
                     int d = Inf;
                     for (int j = ; j < n; ++j)
                         d = min(d, dist(bX[i], bY[i], cX[j], cY[j]));
                     F.add_edge(, i + , d);
                 }
                 for (int i = ; i < k; ++i){
                     int d = Inf;
                     for (int j = ; j < n; ++j)
                         d = min(d, dist(pX[i], pY[i], cX[j], cY[j]));
                     F.add_edge(i +  + m, F.T, d);
                 }
              //   for (int i = 0; i < F.tot; ++i)
                //     printf("%d %d %d\n",F.e[i].y, F.e[i].f, F.e[i].next);
                 return F.maxflow();
         }
 
 };