ECNU 2011 Contest Three For Beginners，我的解题报告

题目链接： http://acm.hust.edu.cn:8080/judge/contest/viewContest.action?cid=1465

A - Number Sequence

模式匹配，KMP 算法。

#include < iostream >
2

#include < cstdio >
3

using namespace std;
5

template < class T >
7

void KMPinit( const T * pat, int patLen, int * flink )

{
8

int j, k; flink[ 0 ] = -1;
9

for( j = 1; j < patLen; ++j )

{
10

k = flink[ j - 1 ];
11

while( ( k != -1 ) && ( pat[ j - 1 ] != pat[ k ] ) )

{ k = flink[ k ]; }
12

flink[ j ] = k + 1;
13

}
14

}
15

template < class T >
16

int KMPmatch( const T * txt, int txtLen, const T * pat, int patLen, const int * flink, int matBegin = 0 )

{
17

int i = matBegin, j = 0;
18

while( ( i < txtLen ) && ( j < patLen ) )

{
19

while( ( j != -1 ) && ( txt[ i ] != pat[ j ] ) )

{ j = flink[ j ]; }
20

++j; ++i;
21

}
22

return ( j >= patLen ? i - patLen : -1 );
23

}
24

const int N = 1000009 ;
26

const int M = 10009 ;
27

int a[ N ], b[ M ], link[ M ], n, m;
29

int main()

{
31

int td, i;
32

scanf( "%d", &td );
33

while ( td-- > 0 )

{
34

scanf( "%d%d", &n, &m );
35

for ( i = 0; i < n; ++i )
36

scanf( "%d", a+i );
37

for ( i = 0; i < m; ++i )
38

scanf( "%d", b+i );
39

KMPinit( b, m, link );
40

i = KMPmatch( a, n, b, m, link, 0 );
41

printf( "%d\n", ( (i==(-1)) ? (-1) : (i+1) ) );
42

}
43

return 0;
44

}
45

B - Big Number

模拟手工笔算就好了，不需要高精度。

#include < iostream >
2

#include < cstdio >
3

using namespace std;
5

const int L = 1009 ;
7

char s[ L ];
9

int main()

{
11

int a, b;
12

char *p;
13

while ( scanf( "%s%d", s, &b ) == 2 )

{
14

a = 0;
15

for ( p = s; *p; ++p )

{
16

a = ( a*10 + (*p) - '0' ) % b;
17

}
18

printf( "%d\n", a );
19

}
20

return 0;
21

}
22

C - A strange lift

最短路径，宽度优先搜索，动态规划，都可以。

我的最短路径 SPFA 算法。

#include < iostream >
2

#include < cstdio >
3

#include < cstring >
4

using namespace std;
6

const int N = 209 ;
8

const int INF = 0x3f3f3f3f ;
9

const int QL = N;
10

int n, a, b, k[ N ];
12

int solve()

{
14

int que[ QL ], inq[ N ], qh, qt, f[ N ], i, j;
15

if ( a == b )

{
16

return 0;
17

}
18

qh = 0;
19

qt = 1;
20

memset( inq, 0, sizeof(inq) );
21

que[ qh ] = a;
22

inq[ a ] = 1;
23

memset( f, 0x3f, sizeof(f) );
24

f[ a ] = 0;
25

while ( qh != qt )

{
26

i = que[ qh ];
27

qh = ( qh + 1 ) % QL;
28

inq[ i ] = 0;
29

j = i + k[ i ];
30

if ( j <= n )

{
31

if ( f[ i ] + 1 < f[ j ] )

{
32

f[ j ] = f[ i ] + 1;
33

if ( !inq[ j ] )

{
34

inq[ j ] = 1;
35

que[ qt ] = j;
36

qt = ( qt + 1 ) % QL;
37

}
38

}
39

}
40

j = i - k[ i ];
41

if ( j >= 1 )

{
42

if ( f[ i ] + 1 < f[ j ] )

{
43

f[ j ] = f[ i ] + 1;
44

if ( !inq[ j ] )

{
45

inq[ j ] = 1;
46

que[ qt ] = j;
47

qt = ( qt + 1 ) % QL;
48

}
49

}
50

}
51

}
52

return ( (f[ b ] == INF) ? (-1) : f[ b ] );
53

}
54

int main()

{
56

int i;
57

for ( ; ; )

{
58

scanf( "%d", &n );
59

if ( n == 0 )

{
60

break;
61

}
62

scanf( "%d%d", &a, &b );
63

for ( i = 1; i <= n; ++i )

{
64

scanf( "%d", k+i );
65

}
66

printf( "%d\n", solve() );
67

}
68

return 0;
69

}
70

D - Intersecting Lines

计算几何入门题，用向量叉积判断直线关系，对于交点，我是用向量推出二元一次方程组，解得交点。

#include < iostream >
2

#include < iomanip >
3

using namespace std;
5

typedef long long Lint;
7

Lint cross( Lint ax, Lint ay, Lint bx, Lint by )

{
9

return ax*by - ay*bx;
10

}
11

int point_in_line( Lint px, Lint py, Lint ax, Lint ay, Lint bx, Lint by )

{
13

return ( cross( ax-bx, ay-by, px-bx, py-by ) == 0 );
14

}
15

int solve( Lint a, Lint b, Lint c, Lint e, Lint f, Lint g, double * x, double * y )

{
17

*y = (double)(c*e-a*g) / (a*f-b*e);
18

*x = (double)(c*f-b*g) / (b*e-a*f);
19

return 1;
20

}
21

#define LINE_ONE 0
23

#define LINE_PAR 1
24

#define LINE_INS 2
25

double ix, iy;
27

int type( Lint ax, Lint ay, Lint bx, Lint by, Lint cx, Lint cy, Lint dx, Lint dy )

{
29

if ( point_in_line(ax,ay,cx,cy,dx,dy) && point_in_line(bx,by,cx,cy,dx,dy) )

{
30

return LINE_ONE;
31

}
32

if ( cross( ax-bx, ay-by, cx-dx, cy-dy ) == 0 )

{
33

return LINE_PAR;
34

}
35

solve( by-ay, ax-bx, ay*(bx-ax)-ax*(by-ay), dy-cy, cx-dx, cy*(dx-cx)-cx*(dy-cy), &ix, &iy );
36

return LINE_INS;
37

}
38

int main()

{
40

int td;
41

Lint ax, ay, bx, by, cx, cy, dx, dy;
42

cout << "INTERSECTING LINES OUTPUT\n";
43

cin >> td;
44

while ( td-- > 0 )

{
45

cin >> ax >> ay >> bx >> by >> cx >> cy >> dx >> dy;
46

switch ( type( ax, ay, bx, by, cx, cy, dx, dy ) )

{
47

case LINE_ONE :
48

cout << "LINE\n";
49

break;
50

case LINE_PAR :
51

cout << "NONE\n";
52

break;
53

case LINE_INS :
54

cout << fixed << setprecision(2) << "POINT " << ix << " " << iy << "\n";
55

break;
56

}
57

}
58

cout << "END OF OUTPUT\n";
59

return 0;
60

}
61

E - Hat’s Words

字符串查找，可以字典树，可以 map 等等。

#include < iostream >
2

#include < string >
3

#include < set >
4

#include < list >
5

using namespace std;
7

int main()

{
9

set< string > dict;
10

list< string > book;
11

string word;
12

int i;
13

while ( cin >> word )

{
14

book.push_back( word );
15

dict.insert( word );
16

}
17

for ( list< string >::const_iterator ite = book.begin(); ite != book.end(); ++ite )

{
18

word = (*ite);
19

for ( i = 1; i+1 < word.length(); ++i )

{
20

if ( (dict.find(word.substr(0,i))!=dict.end()) && (dict.find(word.substr(i))!=dict.end()) )

{
21

cout << word << "\n";
22

break;
23

}
24

}
25

}
26

// system( "pause" );
27

return 0;
28

}
29

F - KILLER

HUST Monthly 2011.04.09的B题，我比赛时没做出来的水题，wbw 告诉我的解法，想想群的知识。

#include < iostream >
2

#include < cstdio >
3

using namespace std;
5

int main()

{
7

int td, a, b, n, ans;
8

scanf( "%d", &td );
9

while ( td-- > 0 )

{
10

scanf( "%d", &n );
11

ans = n;
12

while ( n-- > 0 )

{
13

scanf( "%d%d", &a, &b );
14

if ( a == b )

{
15

--ans;
16

}
17

}
18

printf( "%d\n", ans );
19

}
20

return 0;
21

}
22

G - Conference Call

连接三点的路径中必有一点为三岔口（也许为三点之一），若以此三岔口与三点的最近距离的和为此三岔口的价值，枚举所有三岔口，找出最大的价值，即为答案。补充：赛后和CY讨论，知道枚举所有点即可，不必为三岔口，因为若此点不是三岔口，则价值必定不是最小。

#include < iostream >
2

#include < cstdio >
3

#include < cstring >
4

#include < map >
5

using namespace std;
7

const int N = 10009 ;
9

const int M = 509 ;
10

const int INF = 0x2f2f2f2f ;
11

int disA[ M ], disB[ M ], disC[ M ];
13

int preA[ M ], preB[ M ], preC[ M ];
14

int w[ M ][ M ], m, n, t[ N ];
16

int nosame( int v )

{
18

map< int, int > cnt;
19

map< int, int >::const_iterator ite;
20

int i;
21

cnt[ v ] = 1;
22

for ( i = preA[ v ]; i && preA[ i ]; i = preA[ i ] )

{
23

++cnt[ i ];
24

}
25

for ( i = preB[ v ]; i && preB[ i ]; i = preB[ i ] )

{
26

++cnt[ i ];
27

}
28

for ( i = preC[ v ]; i && preC[ i ]; i = preC[ i ] )

{
29

++cnt[ i ];
30

}
31

for ( ite = cnt.begin(); ite != cnt.end(); ++ite )

{
32

if ( ite->second > 1 )

{
33

return 0;
34

}
35

}
36

return 1;
37

}
38

void spfa( int s, int dis[], int pre[] )

{
40

const int QL = M;
41

int que[ QL ], inq[ M ], qh = 0, qt = 1, i, j;
42

memset( dis, 0x2f, sizeof(int)*M );
44

memset( pre, 0, sizeof(int)*M );
45

memset( inq, 0, sizeof(inq) );
46

dis[ s ] = 0;
47

que[ qh ] = s;
48

inq[ s ] = 1;
49

while ( qh != qt )

{
50

i = que[ qh ];
51

qh = ( qh + 1 ) % QL;
52

inq[ i ] = 0;
53

for ( j = 1; j <= m; ++j )

{
54

if ( dis[ i ] + w[ i ][ j ] < dis[ j ] )

{
55

dis[ j ] = dis[ i ] + w[ i ][ j ];
56

pre[ j ] = i;
57

if ( ! inq[ j ] )

{
58

inq[ j ] = 1;
59

que[ qt ] = j;
60

qt = ( qt + 1 ) % QL;
61

}
62

}
63

}
64

}
65

}
66

// INF <==> no answer
68

int solve( int a, int b, int c )

{
69

int i, ans = INF;
70

spfa( a, disA, preA );
71

spfa( b, disB, preB );
72

spfa( c, disC, preC );
73

for ( i = 1; i <= m; ++i )

{
74

if ( nosame(i) && (disA[i]+disB[i]+disC[i]<ans) )

{
75

ans = disA[ i ] + disB[ i ] + disC[ i ];
76

}
77

}
78

return ans;
79

}
80

int main()

{
82

int lw, i, j, k, a, b, c, cc = 0, cl, ans;
83

while ( scanf( "%d%d%d", &n, &m, &lw ) == 3 )

{
84

printf( "Case #%d\n", ++cc );
85

memset( w, 0x2f, sizeof(w) );
86

for ( i = 1; i <= n; ++i )

{
87

scanf( "%d", t+i );
88

}
89

while ( lw-- > 0 )

{
90

scanf( "%d%d%d", &i, &j, &k );
91

if ( w[ i ][ j ] > k )

{
92

w[ j ][ i ] = w[ i ][ j ] = k;
93

}
94

}
95

for ( i = 1; i <= m; ++i )

{
96

w[ i ][ i ] = 0;
97

}
98

scanf( "%d", &lw );
99

cl = 0;
100

while ( lw-- > 0 )

{
101

scanf( "%d%d%d", &a, &b, &c );
102

ans = solve( t[a], t[b], t[c] );
103

if ( ans == INF )

{
104

printf( "Line %d: Impossible to connect!\n", ++cl );
105

}
106

else

{
107

printf( "Line %d: The minimum cost for this line is %d.\n", ++cl, ans );
108

}
109

}
110

}
111

return 0;
112

}
113

H - How to Type

动态规划，f[ 0 ][ i ] 表示输入前 i 个字符后没有大写锁定的最少击键次数，
f[ 1 ][ i ] 表示输入前 i 个字符后大写锁定的最少击键次数。

#include < iostream >
2

#include < cstdio >
3

#include < cstring >
4

using namespace std;
6

const int L = 109 ;
8

int main()

{
10

char str[ L ];
11

int td, f[ 2 ][ L ], i, tmp;
12

scanf( "%d", &td );
13

while ( td-- > 0 )

{
14

scanf( "%s", str+1 );
15

memset( f, 0x3f, sizeof(f) );
16

f[ 0 ][ 0 ] = 0;
17

for ( i = 1; str[i]; ++i )

{
18

if ( ('a'<=str[i])&&(str[i]<='z') )

{
19

tmp = 1 + f[ 0 ][ i - 1 ];
20

if ( tmp < f[ 0 ][ i ] ) f[ 0 ][ i ] = tmp;
21

tmp = 2 + f[ 1 ][ i - 1 ];
23

if ( tmp < f[ 0 ][ i ] ) f[ 0 ][ i ] = tmp;
24

tmp = 2 + f[ 0 ][ i - 1 ];
26

if ( tmp < f[ 1 ][ i ] ) f[ 1 ][ i ] = tmp;
27

tmp = 2 + f[ 1 ][ i - 1 ];
29

if ( tmp < f[ 1 ][ i ] ) f[ 1 ][ i ] = tmp;
30

}
31

else

{
32

tmp = 2 + f[ 0 ][ i - 1 ];
33

if ( tmp < f[ 0 ][ i ] ) f[ 0 ][ i ] = tmp;
34

tmp = 2 + f[ 1 ][ i - 1 ];
36

if ( tmp < f[ 0 ][ i ] ) f[ 0 ][ i ] = tmp;
37

tmp = 2 + f[ 0 ][ i - 1 ];
39

if ( tmp < f[ 1 ][ i ] ) f[ 1 ][ i ] = tmp;
40

tmp = 1 + f[ 1 ][ i - 1 ];
42

if ( tmp < f[ 1 ][ i ] ) f[ 1 ][ i ] = tmp;
43

}
44

}
45

printf( "%d\n", f[ 0 ][ i-1 ] );
46

}
47

return 0;
48

}
49

比赛时间仓促，自己的做法并非最优。。。

ECNU 2011 Contest Three For Beginners，我的解题报告

ECNU 2011 Contest Three For Beginners，我的解题报告

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