摘要:搜索的过程中用2进制保存状态,通过位运算判断可行性。//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair pii;#defin
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摘要:按位搜索//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair pii;#define pb(a) push(a)#define
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摘要:DP入门题。//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair pii;#define pb(a) push(a)#defi
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摘要:预处理全部q2+p2有20000多个。枚举任意两个的差,并以这个差扩展找到连续的有几个为了避免重复浪费时间,只需判断无法向左扩展的一对数,具体见代码。//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typed
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摘要:迭代加深搜索,预处理9个move.//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair pii;#define pb(a) p
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摘要:太恶心了//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair pii;#define pb(a) push(a)#define
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摘要:按Y轴排序后贪心//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair pii;#define pb(a) push(a)#de
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摘要:#include#include#include#include#include#include#includeusing namespace std;char da[55][55][5];int num[55];int pilenum;void erase(int idx){ for(int i=idx;i3) { if(da[i][num[i]][0]==da[i-3][num[i-3]][0]|| da[i][num[i]][1]==da[i-3][num[i-3]][1]) { ...
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