2017 ACM Arabella Collegiate Programming Contest G. Snake Rana GYM101350G

先算一下总的子矩阵个数

总共最多只有20个点

状压一下然后枚举
容斥一下
把求出来的答案减去包含1个点的子矩阵再加上包含2个点的减去3个点的……

#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <queue>
#include <cstdio>
#include <map>
#include <set>
#include <utility>
#include <stack>
#include <cstring>
#include <cmath>
#include <vector>
#include <ctime>
#include <bitset>
using namespace std;
#define pb push_back
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define ss(str) scanf("%s",str)
#define ans() printf("%d",ans)
#define ansn() printf("%d\n",ans)
#define anss() printf("%d ",ans)
#define lans() printf("%lld",ans)
#define lanss() printf("%lld ",ans)
#define lansn() printf("%lld\n",ans)
#define fansn() printf("%.6f\n",ans)
#define r0(i,n) for(int i=0;i<(n);++i)
#define r1(i,e) for(int i=1;i<=e;++i)
#define rn(i,e) for(int i=e;i>=1;--i)
#define rsz(i,v) for(int i=0;i<(int)v.size();++i)
#define szz(x) ((int)x.size())
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define lowbit(a) (a&(-a))
#define all(a) a.begin(),a.end()
#define pii pair<int,int>
#define pli pair<ll,int>
#define pll pair<ll,ll>
#define mp make_pair
#define lrt rt<<1
#define rrt rt<<1|1
#define X first
#define Y second
#define PI (acos(-1.0))
#define sqr(a) ((a)*(a))
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1000000000+7;
const double eps=1e-9;
const int inf=0x3f3f3f3f;
const ll infl = 10000000000000000;
const int maxn=  200000+10;
const int maxm = 40000+10;
//Pretests passed
int in(int &ret)
{char c;int sgn ;if(c=getchar(),c==EOF)return -1;while(c!='-'&&(c<'0'||c>'9'))c=getchar();sgn = (c=='-')?-1:1;ret = (c=='-')?0:(c-'0');while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');ret *=sgn;return 1;
}pii a[maxn];
pii lu(pii q,pii w)
{int x = min(q.X,w.X);int y = min(q.Y,w.Y);return mp(x,y);
}
pii rd(pii q,pii w)
{int x = max(q.X,w.X);int y = max(q.Y,w.Y);return mp(x,y);
}
int main()
{
#ifdef LOCALfreopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);
#endif // LOCALint t;sd(t);while(t--){int n,m;sdd(n,m);int k;sd(k);r1(i,k){int x , y;sdd(x,y);a[i] = mp(x,y);}int mx = 1<<k;ll ans = (1LL*n*(n+1)/2)*(1LL*m*(m+1)/2);
//        for(int i=1;i<=n;++i)
//        {//            for(int j = 1;j<=m;++j)
//            {//                ll can = (1LL*n-i+1)*(1LL*m-j+1);
//                ans += can;
//            }
//        }for(int i=0;i<mx;++i){int cnt = 0;pii ul = mp(n,m), dr = mp(1,1);for(int j = 0;j<k;++j)if((1<<j)&i)ul = lu(ul,a[j+1]) , dr = rd(dr,a[j+1]),++cnt;if(!cnt)continue;int idx = cnt&1 ? -1:1;ll q = 1LL*ul.X*ul.Y;ll w = (1LL*n - dr.X +1 )*(1LL*m - dr.Y+1);ans += q*w*idx;}lansn();}return 0;
}

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