线段树(3)

hdu 4614  Vases and Flowers

0~n-1的区间，初始值为0，m个操作。
两种操作：
1 X Y 从位置X开始寻找Y个0，如果不足Y个，则寻找尽量多的0，并将他们的值全部修改为1，输出第一个和最后一个修改的1的位置。
2 X Y 输出区间[ X , Y ]内1的个数，并将区间内的1修改为0。

做法：记录区间内1的个数。。关键是第一种操作，ret表示区间[x,n]内0的个数，tmp表示[1,x-1]内0的个数。查询的时候看ls内0的个数是不是大于等于c。

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cmath>
  4 #include<string>
  5 #include<cstring>
  6 #include<algorithm>
  7 #include<set>
  8 using namespace std;
  9 
 10 #define ULL unsigned long long
 11 #define eps 1e-9
 12 #define inf 0x3f3f3f3f
 13 #define ls i << 1
 14 #define rs ls | 1
 15 #define md ((ll + rr) >> 1)
 16 #define lson ll, md, ls
 17 #define rson md + 1, rr, rs
 18 #define LL long long
 19 #define N 50010
 20 #define M 200020
 21 #define mod 1000000007
 22 
 23 int sum[N<<2], down[N<<2];
 24 void build(int ll, int rr, int i){
 25     sum[i] = 0, down[i] = -1;
 26     if(ll == rr) return ;
 27     build(lson), build(rson);
 28 }
 29 void push_down(int i, int ll, int rr){
 30     if(down[i] != -1){
 31         down[ls] = down[rs] = down[i];
 32         sum[ls] = down[i] ? md - ll + 1 : 0;
 33         sum[rs] = down[i] ? rr - md : 0;
 34         down[i] = -1;
 35     }
 36 }
 37 void push_up(int i){
 38     sum[i] = sum[ls] + sum[rs];
 39 }
 40 void update(int l, int r, int v, int ll, int rr, int i){
 41     if(l == ll && r == rr){
 42         down[i] = v;
 43         sum[i] = v ? rr - ll + 1 : 0;
 44         return ;
 45     }
 46     push_down(i, ll, rr);
 47     if(r <= md) update(l, r, v, lson);
 48     else if(l > md) update(l, r, v, rson);
 49     else update(l, md, v, lson), update(md+1, r, v, rson);
 50     push_up(i);
 51 }
 52 int query(int l, int r, int ll, int rr, int i){
 53     if(l == ll && r == rr)
 54         return sum[i];
 55     push_down(i, ll, rr);
 56     int ret;
 57     if(r <= md) ret = query(l, r, lson);
 58     else if(l > md) ret = query(l, r, rson);
 59     else ret = query(l, md, lson) + query(md + 1, r, rson);
 60     push_up(i);
 61     return ret;
 62 }
 63 
 64 int qR(int c, int ll, int rr, int i){
 65     if(ll == rr)
 66         return ll;
 67     push_down(i, ll, rr);
 68     int ret;
 69     int tmp = md - ll + 1 - sum[ls];
 70     if(tmp >= c)
 71         ret = qR(c, lson);
 72     else ret = qR(c - tmp, rson);
 73     push_up(i);
 74     return ret;
 75 }
 76 int main(){
 77     int cas;
 78     //freopen("tt.txt", "r", stdin);
 79     scanf("%d", &cas);
 80     while(cas--){
 81         int n, m;
 82         scanf("%d%d", &n, &m);
 83         build(1, n, 1);
 84         while(m--){
 85             int op, x, y;
 86             scanf("%d%d%d", &op, &x, &y);
 87             if(op == 1){
 88                 x++;
 89                 int ret = n - x + 1 - query(x, n, 1, n, 1);
 90                 int tmp = n - sum[1] - ret;
 91                 if(ret == 0){
 92                     puts("Can not put any one."); continue;
 93                 }
 94                 y = min(y, ret);
 95                 int l = qR(1 + tmp, 1, n, 1);
 96                 int r = qR(y + tmp, 1, n, 1);
 97                 update(l, r, 1, 1, n, 1);
 98                 l--, r--;
 99                 printf("%d %d
", l, r);
100             }
101             else{
102                 x++, y++;
103                 printf("%d
", query(x, y, 1, n, 1));
104                 update(x, y, 0, 1, n, 1);
105             }
106         }
107         puts("");
108     }
109     return 0;
110 }
View Code

 zoj 3299 Fall the Brick

有n个木块，m个板。木块从上往下掉，覆盖的范围是[l, r]，板的高度h不一样，能够接到木块的范围是[L, R]，问每块板上能接多少木块。

做法：离散化，先确定每个区间的id，然后线段树区间更新。最后再做一次query。

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cmath>
  4 #include<string>
  5 #include<cstring>
  6 #include<algorithm>
  7 #include<ctime>
  8 #include<cstdlib>
  9 #include<queue>
 10 using namespace std;
 11 
 12 #define ULL unsigned long long
 13 #define eps 1e-9
 14 #define inf 0x3f3f3f3f
 15 #define ls i << 1
 16 #define rs ls | 1
 17 #define md ((ll + rr) >> 1)
 18 #define lson ll, md, ls
 19 #define rson md + 1, rr, rs
 20 #define LL long long
 21 #define mod 1000000007
 22 #define N 100020
 23 #define M 800020
 24 
 25 LL ans[N], a[N<<2];
 26 int add[N<<4], id[N<<4];
 27 void push_down(int i){
 28     if(id[i]){
 29         id[ls] = id[rs] = id[i];
 30         id[i] = 0;
 31     }
 32     if(add[i]){
 33         add[ls] += add[i], add[rs] += add[i];
 34         add[i] = 0;
 35     }
 36 }
 37 void update(int l, int r, int u, int ll, int rr, int i){
 38     if(l == ll && r == rr){
 39         if(u) id[i] = u;
 40         else add[i]++;
 41         return ;
 42     }
 43     push_down(i);
 44     if(r <= md) update(l, r, u, lson);
 45     else if(l > md) update(l, r, u, rson);
 46     else update(l, md, u, lson), update(md + 1, r, u, rson);
 47 }
 48 
 49 void query(int ll, int rr, int i){
 50     if(ll == rr){
 51         ans[id[i]] += (LL)(a[rr+1] - a[rr]) * add[i];
 52         return;
 53     }
 54     push_down(i);
 55     query(lson), query(rson);
 56 }
 57 struct query{
 58     int L, R, H, id;
 59     void input(int i){
 60         scanf("%d%d%d", &L, &R, &H);
 61         id = i;
 62     }
 63     bool operator < (const query &b) const {
 64         return H < b.H;
 65     }
 66 }q[N];
 67 int L[N], R[N];
 68 int main(){
 69     int n, m;
 70     while(scanf("%d%d", &n, &m) != EOF){
 71         memset(ans, 0, sizeof ans);
 72         memset(id, 0, sizeof id);
 73         memset(add, 0, sizeof add);
 74         int cnt = 0;
 75         for(int i = 1; i <= n; ++i){
 76             scanf("%d%d", &L[i], &R[i]);
 77             a[++cnt] = L[i], a[++cnt] = R[i];
 78         }
 79         for(int i = 1; i <= m; ++i){
 80             q[i].input(i);
 81             a[++cnt] = q[i].L, a[++cnt] = q[i].R;
 82         }
 83         sort(a + 1, a + 1 + cnt);
 84         cnt = unique(a + 1, a + 1 + cnt) - a - 1;
 85         sort(q + 1, q + 1 + m);
 86         for(int i = 1; i <= m; ++i){
 87             int l = lower_bound(a + 1, a + 1 + cnt, q[i].L) - a;
 88             int r = lower_bound(a + 1, a + 1 + cnt, q[i].R) - a;
 89             update(l, r - 1, q[i].id, 1, cnt, 1);
 90         }
 91         for(int i = 1; i <= n; ++i){
 92             int l = lower_bound(a + 1, a + 1 + cnt, L[i]) - a;
 93             int r = lower_bound(a + 1, a + 1 + cnt, R[i]) - a;
 94             update(l, r - 1, 0, 1, cnt, 1);
 95         }
 96         query(1, cnt, 1);
 97         for(int i = 1; i <= m; ++i){
 98             printf("%lld
", ans[i]);
 99         }
100         puts("");
101     }
102     return 0;
103 }
View Code

CF 266E More Queries to Array...

给你一串数，编号从1~n，进行m次操作。（1 <= n , m <= 10 ^ 5 ）

= L R X 区间【L，R】上的数设为X

？L R K 输出  。

做法：对询问进行二项式展开，得到 segma(j = 0, k)[C(k, j) * (1 - l)^(k - j)] * segma(i = l, r)[ai * i^j]。。由于0 <= k <= 5，对于后面部分可以直接开6棵线段树保存。然后前面部分预处理出组合数，询问的时候乘上就好了。

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 #include <cmath>
  6 #include <queue>
  7 #include <stack>
  8 using namespace std;
  9 
 10 #define LL long long
 11 #define eps 1e-8
 12 #define inf 0x3f3f3f3f
 13 #define MP make_pair
 14 #define N 100020
 15 #define M 400020
 16 #pragma comment(linker, "/STACK:1024000000,1024000000")
 17 #define Pi acos(-1.0)
 18 #define mod 1000000007
 19 #define ls (i << 1)
 20 #define rs (ls | 1)
 21 #define md ((ll + rr) >> 1)
 22 #define lson ll, md, ls
 23 #define rson md + 1, rr, rs
 24 
 25 LL c[10][10], s[N][6];
 26 void init(){
 27     c[0][0] = 1;
 28     for(int i = 1; i <= 6; ++i){
 29         c[i][0] = 1;
 30         for(int j = 1; j <= i; ++j)
 31             c[i][j] = c[i-1][j] + c[i-1][j-1];
 32     }
 33     for(int i = 0; i < N; ++i){
 34         for(int j = 0; j < 6; ++j){
 35             s[i][j] = 1;
 36             for(int k = 0; k < j; ++k)
 37                 s[i][j] = s[i][j] * i % mod;
 38         }
 39     }
 40     for(int i = 1; i < N; ++i)
 41         for(int j = 0; j < 6; ++j)
 42             s[i][j] = (s[i][j] + s[i-1][j]) % mod;
 43 }
 44 LL sum[N<<2][6];
 45 int down[N<<2];
 46 void f(int v, int ll, int rr, int i){
 47     down[i] = v;
 48     for(int j = 0; j < 6; ++j)
 49         sum[i][j] = (s[rr][j] - s[ll-1][j] + mod) % mod * v % mod;
 50 }
 51 void push_down(int i, int ll, int rr){
 52     if(~down[i]){
 53         f(down[i], lson);
 54         f(down[i], rson);
 55         down[i] = -1;
 56     }
 57 }
 58 void push_up(int i){
 59     for(int j = 0; j < 6; ++j)
 60         sum[i][j] = (sum[ls][j] + sum[rs][j]) % mod;
 61 }
 62 void build(int ll, int rr, int i){
 63     down[i] = -1;
 64     if(ll == rr){
 65         int v;
 66         scanf("%d", &v);
 67         for(int j = 0; j < 6; ++j)
 68             sum[i][j] = (s[rr][j] - s[rr-1][j] + mod) % mod * v % mod;
 69         return ;
 70     }
 71     build(lson), build(rson);
 72     push_up(i);
 73 }
 74 
 75 void update(int l, int r, int v, int ll, int rr, int i){
 76     if(l == ll && r == rr){
 77         f(v, ll, rr, i);
 78         return ;
 79     }
 80     push_down(i, ll, rr);
 81     if(r <= md) update(l, r, v, lson);
 82     else if(l > md) update(l, r, v, rson);
 83     else update(l, md, v, lson), update(md + 1, r, v, rson);
 84     push_up(i);
 85 }
 86 LL query(int l, int r, int p, int v, int ll, int rr, int i){
 87     if(l == ll && r == rr){
 88         LL ret = 0, tmp = 1;
 89         for(int j = v; j >= 0; --j){
 90             //printf("%lld
", sum[i][j]);
 91             ret = (ret + sum[i][j] * c[v][j] % mod * tmp % mod + mod) % mod;
 92             tmp = tmp * (1 - p) % mod;
 93            // printf("%lld
", ret);
 94         }
 95         return ret;
 96     }
 97     push_down(i, ll, rr);
 98     LL ret;
 99     if(r <= md) ret = query(l, r, p, v, lson);
100     else if(l > md) ret = query(l, r, p, v, rson);
101     else ret = (query(l, md, p, v, lson) + query(md + 1, r, p, v, rson)) % mod;
102     push_up(i);
103     return ret;
104 }
105 int main(){
106     init();
107     int n, m;
108     scanf("%d%d", &n, &m);
109     build(1, n, 1);
110     while(m--){
111         int L, R, k;
112         char ch[5];
113         scanf("%s%d%d%d", &ch, &L, &R, &k);
114         if(ch[0] == '=')
115             update(L, R, k, 1, n, 1);
116         else printf("%I64d
", query(L, R, L, k, 1, n, 1));
117     }
118     return 0;
119 }
View Code

hdu 4417 Super Mario

长为n的序列，q次询问，每次输出[l, r]内小于h的个数

做法：离散化，将询问排序。线段树搞一下。（将初始序列排序，询问再排序，用树状数组也能搞，而且更快）

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cmath>
  4 #include<string>
  5 #include<cstring>
  6 #include<algorithm>
  7 #include<set>
  8 #include<queue>
  9 #include <bitset>
 10 using namespace std;
 11 
 12 #define ULL unsigned long long
 13 #define eps 1e-9
 14 #define inf 0x3f3f3f3f
 15 #define ls i << 1
 16 #define rs ls | 1
 17 #define md ((ll + rr) >> 1)
 18 #define lson ll, md, ls
 19 #define rson md + 1, rr, rs
 20 #define LL long long
 21 #define N 100100
 22 #define M 200200
 23 #define mod 2015
 24 #define MP make_pair
 25 #define PB push_back
 26 
 27 int sum[N<<4];
 28 void update(int p, int ll, int rr, int i){
 29     if(ll == rr){
 30         sum[i]++;
 31         return ;
 32     }
 33     if(p <= md) update(p, lson);
 34     else update(p, rson);
 35     sum[i] = sum[ls] + sum[rs];
 36 }
 37 int query(int l, int r, int ll, int rr, int i){
 38     if(l == ll && r == rr)
 39         return sum[i];
 40     if(r <= md) return query(l, r, lson);
 41     if(l > md) return query(l, r, rson);
 42     return query(l, md, lson) + query(md + 1, r, rson);
 43 }
 44 struct node{
 45     int p, h, id;
 46     node(int p = 0, int h = 0, int id = 0) : p(p), h(h), id(id) {}
 47     bool operator < (const node &b) const{
 48         return p < b.p;
 49     }
 50 }L[N], R[N];
 51 int cnt, a[N], b[N<<1], qL[N], qR[N];
 52 int Hash(int v){
 53     return (int)(lower_bound(b + 1, b + 1 + cnt, v) - b);
 54 }
 55 void debug(){
 56     for(int i = 1; i <= cnt; ++i)
 57         printf("%d ", b[i]);
 58     cout << endl;
 59 }
 60 int main(){
 61     int cas, kk = 0;
 62     scanf("%d", &cas);
 63     while(cas--){
 64         int n, m;
 65         scanf("%d%d", &n, &m);
 66         cnt = 0;
 67         for(int i = 1; i <= n; ++i){
 68             scanf("%d", &a[i]);
 69             b[++cnt] = a[i];
 70         }
 71         for(int i = 1; i <= m; ++i){
 72             int l, r, h;
 73             scanf("%d%d%d", &l, &r, &h);
 74             l++, r++;
 75             L[i] = node(l, h, i);
 76             R[i] = node(r, h, i);
 77             b[++cnt] = h;
 78         }
 79         sort(b + 1, b + 1 + cnt);
 80         cnt = unique(b + 1, b + 1 + cnt) - b - 1;
 81         sort(L + 1, L + 1 + m);
 82         sort(R + 1, R + 1 + m);
 83     //    debug();
 84         memset(sum, 0, sizeof sum);
 85         int j = 1, k = 1;
 86         for(int i = 1; i <= n; ++i){
 87             while(L[j].p == i){
 88                 int p = Hash(L[j].h);
 89             //    printf("%d
", p);
 90                 qL[L[j].id] = query(1, p, 1, cnt, 1);
 91                 j++;
 92             }
 93             int p = Hash(a[i]);
 94             update(p, 1, cnt, 1);
 95             while(R[k].p == i){
 96                 int p = Hash(R[k].h);
 97                 qR[R[k].id] = query(1, p, 1, cnt, 1);
 98                 k++;
 99             }
100         }
101         printf("Case %d:
", ++kk);
102         for(int i = 1; i <= m; ++i){
103             printf("%d
", qR[i] - qL[i]);
104         }
105     }
106     return 0;
107 }
View Code
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原文地址：https://www.cnblogs.com/LJ-blog/p/4749242.html