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 }
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 }
