Codeforces Round #149 (Div. 2)

http://codeforces.com/contest/242 

C题：给出一些线段，起点和终点，只能在线段上走，线段范围上限为1e9，但是线段的总长不超过1e5.

做法：bfs+map

View Code

 1 /*

 2 Author:Zhaofa Fang

 3 Lang:C++

 4 */

 5 #include <cstdio>

 6 #include <cstdlib>

 7 #include <sstream>

 8 #include <iostream>

 9 #include <cmath>

10 #include <cstring>

11 #include <algorithm>

12 #include <string>

13 #include <utility>

14 #include <vector>

15 #include <queue>

16 #include <stack>

17 #include <map>

18 #include <set>

19 using namespace std;

20 

21 typedef long long ll;

22 #define DEBUG(x) cout<< #x << ':' << x << endl

23 #define REP(i,n) for(int i=0;i < (n);i++)

24 #define FOR(i,s,t) for(int i = (s);i <= (t);i++)

25 #define PII pair<int,int>

26 #define PB push_back

27 #define MP make_pair

28 #define ft first

29 #define sd second

30 #define lowbit(x) (x&(-x))

31 #define INF (1<<30)

32 

33 PII point;

34 map<PII,int> S,S1;

35 int X0,Y0,X1,Y1;

36 int bfs(void)

37 {

38     S1.clear();

39     queue<PII> Q;

40     Q.push(MP(X0,Y0));

41     S[MP(X0,Y0)] = 0;

42     while(!Q.empty())

43     {

44         int xx = Q.front().ft;

45         int yy = Q.front().sd;

46         Q.pop();

47         FOR(i,-1,1)FOR(j,-1,1)

48         {

49             if(!i && !j)continue;

50             PII next = MP(xx + i,yy + j);

51             if(!S.count(next) || S1.count(next))continue;

52             Q.push(next);

53             S1[next] = S1[MP(xx,yy)] + 1;

54             if(next.ft == X1 && next.sd == Y1)return S1[next];

55         }

56     }

57     return -1;

58 }

59 int main()

60 {

61     //freopen("in","r",stdin);

62     while(cin>>X0>>Y0>>X1>>Y1)

63     {

64         int n;

65         cin>>n;

66         int r,a,b;

67         S.clear();

68         REP(i,n)

69         {

70             cin>>r>>a>>b;

71             FOR(i,a,b)

72             {

73                 S[MP(r,i)]++;

74             }

75         }

76         printf("%d\n",bfs());

77     }

78     return 0;

79 }

 

D题：每一个节点press一次就会+1，其相邻节点也会+1，每一个节点只能press一次，初始每个节点bi都为0，问press哪些节点

使得最后bi != ai。

做法：如果bi == ai就press，可以用bfs直到所有bi != ai。因为bi只会增加，所以一定有解，复杂度为O(n)。

View Code

 1 /*

 2 Author:Zhaofa Fang

 3 Lang:C++

 4 */

 5 #include <cstdio>

 6 #include <cstdlib>

 7 #include <sstream>

 8 #include <iostream>

 9 #include <cmath>

10 #include <cstring>

11 #include <algorithm>

12 #include <string>

13 #include <utility>

14 #include <vector>

15 #include <queue>

16 #include <stack>

17 #include <map>

18 #include <set>

19 using namespace std;

20 

21 typedef long long ll;

22 #define DEBUG(x) cout<< #x << ':' << x << endl

23 #define REP(i,n) for(int i=0;i < (n);i++)

24 #define FOR(i,s,t) for(int i = (s);i <= (t);i++)

25 #define PII pair<int,int>

26 #define PB push_back

27 #define MP make_pair

28 #define FI first

29 #define SE second

30 #define lowbit(x) (x&(-x))

31 #define INF (1<<30)

32 

33 vector<int> vec[100005];

34 int X[100005];

35 int a[100005],b[100005];

36 int main()

37 {

38     //freopen("in","r",stdin);

39     int n,m;

40     while(cin>>n>>m)

41     {

42         queue<int> Q;

43         int k = 0;

44         memset(b,0,sizeof(b));

45         while(m --)

46         {

47             int u,v;

48             cin>>u>>v;

49             vec[u].PB(v);

50             vec[v].PB(u);

51         }

52         FOR(i,1,n)

53         {

54             cin>>a[i];

55             if(a[i] == 0)

56             {

57                 Q.push(i);b[i]++;X[k++]=i;

58                 REP(j,vec[i].size())b[vec[i][j]] ++;

59             }

60         }

61         while(!Q.empty())

62         {

63             int x = Q.front();Q.pop();

64             REP(i,vec[x].size())

65             {

66                 int v = vec[x][i];

67                 //DEBUG(v);

68                 //DEBUG(a[v]);DEBUG(b[v]);

69                 if(b[v] == a[v])

70                 {

71                     Q.push(v);

72                     b[v] ++;

73                     X[k++] = v;

74                     REP(j,vec[v].size())b[vec[v][j]]++;

75                 }

76             }

77         }

78         printf("%d\n",k);

79         REP(i,k)printf("%d ",X[i]);

80         puts("");

81     }

82     return 0;

83 }

 

E题不错

1）区间求和。

2）区间更新，每个数异或X。

做法：把每个数按二进制拆位，故可以建20棵线段树。

View Code

  1 /*

  2 Author:Zhaofa Fang

  3 Lang:C++

  4 */

  5 #include <cstdio>

  6 #include <cstdlib>

  7 #include <sstream>

  8 #include <iostream>

  9 #include <cmath>

 10 #include <cstring>

 11 #include <algorithm>

 12 #include <string>

 13 #include <utility>

 14 #include <vector>

 15 #include <queue>

 16 #include <stack>

 17 #include <map>

 18 #include <set>

 19 using namespace std;

 20 

 21 typedef long long ll;

 22 #define DEBUG(x) cout<< #x << ':' << x << endl

 23 #define REP(i,n) for(int i=0;i < (n);i++)

 24 #define FOR(i,s,t) for(int i = (s);i <= (t);i++)

 25 #define PII pair<int,int>

 26 #define PB push_back

 27 #define MP make_pair

 28 #define FI first

 29 #define SE second

 30 #define lowbit(x) (x&(-x))

 31 #define INF (1<<30)

 32 

 33 #define lson l , m , rt<<1

 34 #define rson m + 1 , r , rt<<1|1

 35 

 36 const int maxn = 100111;

 37 int col[maxn<<2];

 38 ll sum[maxn<<2][22];

 39 

 40 void PushUp(int rt)

 41 {

 42     REP(i,21)

 43     sum[rt][i] = sum[rt<<1][i] + sum[rt<<1|1][i];

 44 }

 45 void PushDown(int rt,int m)

 46 {

 47     if(col[rt])

 48     {

 49         col[rt<<1] ^= col[rt];

 50         col[rt<<1|1] ^= col[rt];

 51         REP(i,21)

 52         {

 53             if(col[rt]&1)

 54             {

 55                 sum[rt<<1][i] = m - (m>>1) - sum[rt<<1][i];

 56                 sum[rt<<1|1][i] = (m>>1) - sum[rt<<1|1][i];

 57             }

 58             col[rt] >>= 1;

 59         }

 60     }

 61 }

 62 void build(int l,int r,int rt)

 63 {

 64     REP(i,21)sum[rt][i] = 0;

 65     col[rt] = 0;

 66     if(l == r)

 67     {

 68         int A;

 69         scanf("%I64d",&A);

 70         REP(i,21)

 71         {

 72             sum[rt][i] = (A&1);

 73             A >>= 1;

 74         }

 75         return ;

 76     }

 77     int m = (l + r) >> 1;

 78     build(lson);

 79     build(rson);

 80     PushUp(rt);

 81 }

 82 void updata(int L,int R,int X,int l,int r,int rt)

 83 {

 84     if(L <= l && r <= R)

 85     {

 86         col[rt] ^= X;

 87         REP(i,21)

 88         {

 89             if(X&1)sum[rt][i] = r - l + 1 - sum[rt][i];//0变1,1变0

 90             X>>=1;

 91         }

 92         return ;

 93     }

 94     PushDown(rt,r-l+1);

 95     int m = (l + r) >> 1;

 96     if(L <= m)updata(L,R,X,lson);

 97     if(m < R)updata(L,R,X,rson);

 98     PushUp(rt);

 99 }

100 ll query(int L,int R,int l,int r,int rt)

101 {

102     if(L <= l && r <= R)

103     {

104         ll tmp = 0;

105         REP(i,21)

106         {

107             tmp += sum[rt][i]*(1LL<<i);

108         }

109         return tmp;

110     }

111     PushDown(rt,r-l+1);

112     int m = (l + r) >> 1;

113     ll ret = 0;

114     if(L <= m)ret += query(L,R,lson);

115     if(m < R)ret += query(L,R,rson);

116     return ret;

117 }

118 int main()

119 {

120     //freopen("in","r",stdin);

121     int n;

122     while(~scanf("%d",&n))

123     {

124         build(1,n,1);

125         int m;

126         scanf("%d",&m);

127         int op,l,r,x;

128         while(m--)

129         {

130             scanf("%d",&op);

131             if(op == 1)

132             {

133                 scanf("%d%d",&l,&r);

134                 printf("%I64d\n",query(l,r,1,n,1));

135             }

136             else

137             {

138                 scanf("%d%d%d",&l,&r,&x);

139                 updata(l,r,x,1,n,1);

140             }

141         }

142     }

143     return 0;

144 }