hdu 3397 线段树

题意： Change operations:
0 a b change all characters into '0's in [a , b]
1 a b change all characters into '1's in [a , b]
2 a b change all '0's into '1's and change all '1's into '0's in [a, b]
Output operations:
3 a b output the number of '1's in [a, b]
4 a b output the length of the longest continuous '1' string in [a , b]

Sample Input
1
10 10
0 0 0 1 1 0 1 0 1 1
1 0 2
3 0 5
2 2 2
4 0 4
0 3 6
2 3 7
4 2 8
1 0 5
0 5 6
3 3 9

Sample Output

5
2
6
5

嗯嗯

2015-07-26:专题复习

  1 #include<cstdio>

  2 #include<iostream>

  3 #include<algorithm>

  4 #include<cstring>

  5 #include<cmath>

  6 #include<queue>

  7 #include<map>

  8 using namespace std;

  9 #define MOD 1000000007

 10 const int INF=0x3f3f3f3f;

 11 const double eps=1e-5;

 12 #define cl(a) memset(a,0,sizeof(a))

 13 #define ts printf("*****\n");

 14 #define lson l,mid,rt<<1

 15 #define rson mid+1,r,rt<<1|1

 16 #define root 1,n,1

 17 #define mid ((l+r)>>1)

 18 const int MAXN=100010;

 19 int n,m,t,Min,tt;

 20 int msum1[MAXN<<2],col[MAXN<<2],lsum1[MAXN<<2],rsum1[MAXN<<2];

 21 int msum0[MAXN<<2],lsum0[MAXN<<2],rsum0[MAXN<<2],xxor[MAXN<<2];

 22 int sum1[MAXN<<2],sum0[MAXN<<2];

 23 void fun1(int rt,int m,int val)     //改变区间内0/1的变化

 24 {

 25     msum1[rt]=lsum1[rt]=rsum1[rt]=sum1[rt]=val?m:0;

 26     msum0[rt]=lsum0[rt]=rsum0[rt]=sum0[rt]=val?0:m;

 27 }

 28 void fun2(int rt)

 29 {

 30     swap(msum1[rt],msum0[rt]);

 31     swap(lsum1[rt],lsum0[rt]);

 32     swap(rsum1[rt],rsum0[rt]);

 33     swap(sum1[rt],sum0[rt]);

 34 }

 35 

 36 void pushup(int rt,int m)

 37 {

 38     sum1[rt]=sum1[rt<<1]+sum1[rt<<1|1];

 39     lsum1[rt]=lsum1[rt<<1];

 40     rsum1[rt]=rsum1[rt<<1|1];

 41     if(lsum1[rt]==(m-(m>>1)))    lsum1[rt]+=lsum1[rt<<1|1];

 42     if(rsum1[rt]==(m>>1))    rsum1[rt]+=rsum1[rt<<1];

 43     msum1[rt]=max(lsum1[rt<<1|1]+rsum1[rt<<1],max(msum1[rt<<1],msum1[rt<<1|1]));

 44 

 45     sum0[rt]=sum0[rt<<1]+sum0[rt<<1|1];

 46     lsum0[rt]=lsum0[rt<<1];

 47     rsum0[rt]=rsum0[rt<<1|1];

 48     if(lsum0[rt]==(m-(m>>1)))    lsum0[rt]+=lsum0[rt<<1|1];

 49     if(rsum0[rt]==(m>>1))    rsum0[rt]+=rsum0[rt<<1];

 50     msum0[rt]=max(lsum0[rt<<1|1]+rsum0[rt<<1],max(msum0[rt<<1],msum0[rt<<1|1]));

 51 }

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

 53 {

 54     col[rt]=-1;

 55     xxor[rt]=0;

 56     if(l==r)

 57     {

 58         int num;

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

 60         msum1[rt]=lsum1[rt]=rsum1[rt]=sum1[rt]=num?1:0;

 61         msum0[rt]=lsum0[rt]=rsum0[rt]=sum0[rt]=num?0:1;

 62         return;

 63     }

 64     build(lson);

 65     build(rson);

 66     pushup(rt,r-l+1);

 67 }

 68 void pushdown(int rt,int m)

 69 {

 70     if(col[rt]!=-1)

 71     {

 72         xxor[rt<<1]=xxor[rt<<1|1]=0;

 73         col[rt<<1]=col[rt<<1|1]=col[rt];

 74         fun1(rt<<1,m-(m>>1),col[rt]);

 75         fun1(rt<<1|1,m>>1,col[rt]);

 76         col[rt]=-1;

 77     }

 78     else if(xxor[rt])

 79     {

 80         if(col[rt<<1]!=-1)

 81         {

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

 83             fun1(rt<<1,m-(m>>1),col[rt<<1]);

 84         }

 85         else

 86         {

 87             xxor[rt<<1]^=1;

 88             fun2(rt<<1);                //最大连续1的变化有点麻烦，必须要重新开一个最大连续0的变量

 89         }

 90         if(col[rt<<1|1]!=-1)

 91         {

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

 93             fun1(rt<<1|1,(m>>1),col[rt<<1|1]);

 94         }

 95         else

 96         {

 97             xxor[rt<<1|1]^=1;

 98             fun2(rt<<1|1);

 99         }

100         xxor[rt]=0;

101     }

102 }

103 void update(int L,int R,int val,int l,int r,int rt)

104 {

105     if(l>=L&&r<=R)

106     {

107         if(val<=1)

108         {

109             xxor[rt]=0;

110             col[rt]=val;

111             fun1(rt,r-l+1,val);

112         }

113         else

114         {

115             if(col[rt]!=-1)

116             {

117                 col[rt]^=1;

118                 fun1(rt,r-l+1,col[rt]);

119             }

120             else

121             {

122                 xxor[rt]^=1;

123                 fun2(rt);

124             }

125         }

126         return;

127     }

128     pushdown(rt,r-l+1);

129     if(L<=mid) update(L,R,val,lson);

130     if(R>mid) update(L,R,val,rson);

131     pushup(rt,r-l+1);

132 }

133 int query1(int L,int R,int l,int r,int rt)  //查询区间内1的个数

134 {

135     if(l>=L&&r<=R)

136     {

137         return sum1[rt];

138     }

139     pushdown(rt,r-l+1);

140     int ans=0;

141     if(L<=mid) ans+=query1(L,R,lson);

142     if(R>mid) ans+=query1(L,R,rson);

143     return ans;

144 }

145 int query2(int L,int R,int l,int r,int rt)  //查询区间内最长连续1的个数

146 {

147     if(l>=L&&r<=R)

148     {

149         return msum1[rt];

150     }

151     pushdown(rt,r-l+1);

152 

153     if(R<=mid) return query2(L,R,lson);

154     if(L>mid) return query2(L,R,rson);

155     int a=query2(L,R,lson);

156     int b=query2(L,R,rson);

157     a=max(a,b);

158     b=min(mid-L+1,rsum1[rt<<1])+min(R-mid,lsum1[rt<<1|1]);

159     return max(a,b);

160 }

161 int main()

162 {

163     #ifndef ONLINE_JUDGE

164     freopen("1.in","r",stdin);

165     #endif

166     int t,n,m,op,a,b;

167     scanf("%d",&t);

168     while(t--)

169     {

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

171         build(root);

172         while(m--)

173         {

174             scanf("%d%d%d",&op,&a,&b);

175             a++,b++;

176             if(op<=2) update(a,b,op,root);

177             else

178             {

179                 if(op==3) printf("%d\n",query1(a,b,root));

180                 else printf("%d\n",query2(a,b,root));

181             }

182         }

183     }

184     return 0;

185 }

View Code

 

 

  1 #include<cstdio>

  2 #include<cstring>

  3 #include<algorithm>

  4 using namespace std;

  5 const int maxn = 100010;

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

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

  8 int num0[maxn<<2],num1[maxn<<2];

  9 int lb0[maxn<<2],lb1[maxn<<2];

 10 int rb0[maxn<<2],rb1[maxn<<2];

 11 int mx1[maxn<<2],mx0[maxn<<2];

 12 int col[maxn<<2];

 13 int xxor[maxn<<2];

 14 int num;

 15 int max(int a,int b){

 16     return a>b?a:b;

 17 }

 18 int min(int a,int b){

 19     return a<b?a:b;

 20 }

 21 void pushup(int rt,int m)

 22 {

 23     lb0[rt]=lb0[rt<<1];rb0[rt]=rb0[rt<<1|1];

 24     if(lb0[rt]==m-(m>>1)) lb0[rt]+=lb0[rt<<1|1];

 25     if(rb0[rt]==(m>>1))  rb0[rt]+=rb0[rt<<1];

 26 

 27     lb1[rt]=lb1[rt<<1];rb1[rt]=rb1[rt<<1|1];

 28     if(lb1[rt]==m-(m>>1)) lb1[rt]+=lb1[rt<<1|1];

 29     if(rb1[rt]==(m>>1))  rb1[rt]+=rb1[rt<<1];

 30 

 31     num0[rt]=num0[rt<<1]+num0[rt<<1|1];

 32     num1[rt]=num1[rt<<1]+num1[rt<<1|1];

 33 

 34     mx1[rt]=max(mx1[rt<<1],mx1[rt<<1|1]);

 35     mx1[rt]=max(mx1[rt],rb1[rt<<1]+lb1[rt<<1|1]);

 36 

 37     mx0[rt]=max(mx0[rt<<1],mx0[rt<<1|1]);

 38     mx0[rt]=max(mx0[rt],rb0[rt<<1]+lb0[rt<<1|1]);

 39 }

 40 void build(int l,int r,int rt){

 41     col[rt]=-1; xxor[rt]=0;

 42     if(l==r){

 43         scanf("%d",&num);

 44         num0[rt]=lb0[rt]=rb0[rt]=mx0[rt]=(num?0:1);

 45         num1[rt]=lb1[rt]=rb1[rt]=mx1[rt]=(num?1:0);

 46         return ;

 47     }

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

 49     build(lson);

 50     build(rson);

 51     pushup(rt,r-l+1);

 52 }

 53 void init(int rt,int m,int cmd){

 54    lb0[rt]=rb0[rt]=mx0[rt]=num0[rt]=m*(1-cmd);

 55    lb1[rt]=rb1[rt]=mx1[rt]=num1[rt]=m*cmd;

 56 }

 57 void change(int rt){

 58     swap(lb0[rt],lb1[rt]);

 59     swap(rb0[rt],rb1[rt]);

 60     swap(mx0[rt],mx1[rt]);

 61     swap(num0[rt],num1[rt]);

 62 }

 63 void pushdown(int rt,int m){

 64     if(col[rt]!=-1){

 65          col[rt<<1]=col[rt<<1|1]=col[rt];

 66          xxor[rt<<1]=xxor[rt<<1|1]=0;

 67          init(rt<<1,m-(m>>1),col[rt]);

 68          init(rt<<1|1,(m>>1),col[rt]);

 69          col[rt]=-1;

 70     }

 71     else if(xxor[rt]){

 72         if(col[rt<<1]!=-1){

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

 74             init(rt<<1,m-(m>>1),col[rt<<1]);

 75         }

 76         else {

 77             xxor[rt<<1]^=1;

 78             change(rt<<1);

 79         }

 80         if(col[rt<<1|1]!=-1){

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

 82             init(rt<<1|1,(m>>1),col[rt<<1|1]);

 83         }

 84         else {

 85             xxor[rt<<1|1]^=1;

 86             change(rt<<1|1);

 87         }

 88         xxor[rt]=0;

 89     }

 90 }

 91 void update(int L,int R,int cmd,int l,int r,int rt){

 92     if(L<=l&&r<=R){

 93         if(cmd<=1){

 94             xxor[rt]=0;//取消异或标记

 95             col[rt]=cmd;//真个区间被完全覆盖为cmd

 96             init(rt,r-l+1,cmd);

 97         }

 98         else {

 99             if(col[rt]!=-1){//被完全覆盖，全是0或者全是1，相当于取反

100                 col[rt]^=1;

101                 init(rt,r-l+1,col[rt]);

102             }

103             else {

104                 xxor[rt]^=1;

105                 change(rt);

106             }

107         }

108         return ;

109     }

110     pushdown(rt,r-l+1);

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

112     if(L<=m) update(L,R,cmd,lson);

113     if(R>m) update(L,R,cmd,rson);

114     pushup(rt,r-l+1);

115 }

116 int query1(int L,int R,int l,int r,int rt){

117     if(L<=l&&r<=R)  return num1[rt];

118     pushdown(rt,r-l+1);

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

120     int ans=0;

121     if(L<=m) ans+=query1(L,R,lson);

122     if(R>m) ans+=query1(L,R,rson);

123     return ans;

124 }

125 int  query2(int L,int R,int l,int r,int rt){

126     if(L<=l&&r<=R) {

127         return mx1[rt];

128     }

129     pushdown(rt,r-l+1);

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

131     if(R<=m) return query2(L,R,lson);

132     if(L>m) return query2(L,R,rson);

133     int a=query2(L,R,lson);

134     int b=query2(L,R,rson);

135     a=a>b?a:b;

136     b=min(m-L+1,rb1[rt<<1])+min(R-m,lb1[rt<<1|1]);

137     return a>b?a:b;

138 }

139 int main(){

140     int t,n,m,op,a,b;

141     scanf("%d",&t);

142     while(t--){

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

144         build(0,n-1,1);

145         while(m--){

146             scanf("%d%d%d",&op,&a,&b);

147             if(op<=2) update(a,b,op,0,n-1,1);

148             else {

149                 if(op==3) printf("%d\n",query1(a,b,0,n-1,1));

150                 else printf("%d\n",query2(a,b,0,n-1,1));

151             }

152         }

153     }

154     return 0;

155 }