pku3256可以用這個(gè)函數(shù)
Floyd-Warshall
An alternative solution is to use the Floyd-Warshall transitive-closure algorith, This is an extremely simple triple-loop that yields the transitive closure - that is, a table indicating whether there is a route from each field to each other field. If you are not familiar with this algorithm, you should look it up on the Internet. It is a handy tool in contests simply because it is very compact.
構(gòu)造函數(shù)
bitset<n> b;
b有n位，每位都為0.參數(shù)n可以為一個(gè)表達(dá)式.
如bitset<5> b0;則"b0"為"00000";
bitset<n> b(unsigned long u);
b有n位,并用u賦值;如果u超過n位,則頂端被截除
如:bitset<5>b0(5);則"b0"為"00101";
bitset<n> b(string s);
b是string對(duì)象s中含有的位串的副本
string bitval ( "10011" );
bitset<5> b0 ( bitval4 );
則"b0"為"10011";
bitset<n> b(s, pos);
b是s中從位置pos開始位的副本,前面的多余位自動(dòng)填充0;
string bitval ("01011010");
bitset<10> b0 ( bitval5, 3 );
則"b0" 為 "0000011010";
bitset<n> b(s, pos, num);
b是s中從位置pos開始的num個(gè)位的副本,如果num<n,則前面的空位自動(dòng)填充0;
string bitval ("11110011011");
bitset<6> b0 ( bitval5, 3, 6 );
則"b0" 為 "100110";
os << b
把b中的位集輸出到os流
os >>b
輸入到b中,如"cin>>b",如果輸入的不是0或1的字符,只取該字符前面的二進(jìn)制位.
bool any( )
是否存在置為1的二進(jìn)制位？和none()相反
bool none( )
是否不存在置為1的二進(jìn)制位,即全部為0？和any()相反.
size_t count( )
二進(jìn)制位為1的個(gè)數(shù).
size_t size( )
二進(jìn)制位的個(gè)數(shù)
flip()
把所有二進(jìn)制位逐位取反
flip(size_t pos)
把在pos處的二進(jìn)制位取反
bool operator[]( size_type _Pos )
獲取在pos處的二進(jìn)制位
set()
把所有二進(jìn)制位都置為1
set(pos)
把在pos處的二進(jìn)制位置為1
reset()
把所有二進(jìn)制位都置為0
reset(pos)
把在pos處的二進(jìn)制位置為0
test(size_t pos)
在pos處的二進(jìn)制位是否為1？
unsigned long to_ulong( )
用同樣的二進(jìn)制位返回一個(gè)unsigned long值
string to_string ()
返回對(duì)應(yīng)的字符串.
詳細(xì)請(qǐng)翻閱msdn.