POJ 3667 Hotel 线段树处理区间信息
题目大意：
Hotel里有N个房间，初始全为空，有两种操作
第一种找到序列中最左端的，满足连续1的个数>=d个的位置，并把这些1变成0
第二种是区间修改L,R变成1
题目分析：
就是区间合并问题，不过有些麻烦
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 #define maxn 50010 7 #define lson l, m, rt << 1 8 #define rson m + 1, r, rt << 1 | 1 9 using namespace std; 10 struct SegTree{ 11 int l,r; 12 int mi,lsum,rsum,sum,res,p; 13 }tr[maxn << 2]; 14 int col[maxn << 2]; 15 int n,Q; 16 inline void pushup(int rt){ 17 tr[rt].sum = tr[rt << 1].sum + tr[rt << 1 | 1].sum; 18 tr[rt].lsum = tr[rt << 1].lsum + (tr[rt << 1].r - tr[rt << 1].l + 1 == tr[rt << 1].sum ? tr[rt << 1 | 1].lsum : 0); 19 tr[rt].rsum = tr[rt << 1 | 1].rsum + (tr[rt << 1 | 1].r - tr[rt << 1 | 1].l + 1 == tr[rt << 1 | 1].sum ? tr[rt << 1].rsum : 0); 20 tr[rt].mi = max(max(tr[rt << 1].mi, tr[rt << 1 | 1].mi), tr[rt << 1].rsum + tr[rt << 1 | 1].lsum); 21 tr[rt].res = tr[rt << 1].rsum + tr[rt << 1 | 1].lsum; 22 tr[rt].p = tr[rt << 1].r - tr[rt << 1].rsum + 1; 23 } 24 inline void pushdown(int rt){ 25 if(col[rt] != -1){ 26 col[rt << 1] = col[rt << 1 | 1] = col[rt]; 27 tr[rt << 1].lsum = tr[rt << 1].rsum = tr[rt << 1].sum = tr[rt << 1].mi = tr[rt << 1].res = (tr[rt << 1].r - tr[rt << 1].l + 1) * col[rt]; 28 tr[rt << 1].p = tr[rt << 1].l; 29 tr[rt << 1 | 1].lsum = tr[rt << 1 | 1].rsum = tr[rt << 1 | 1].sum = tr[rt << 1 | 1].mi = tr[rt << 1 | 1].res = (tr[rt << 1 | 1].r - tr[rt << 1 | 1].l + 1) * col[rt]; 30 tr[rt << 1 | 1].p = tr[rt << 1 | 1].l; 31 col[rt] = -1; 32 } 33 } 34 inline void build(int l,int r,int rt){ 35 tr[rt].l = l, tr[rt].r = r; 36 tr[rt].lsum = tr[rt].rsum = tr[rt].sum = tr[rt].sum = 1; 37 col[rt] = -1; 38 if(l == r){ 39 return; 40 } 41 int m = (l + r) >> 1; 42 build(lson); 43 build(rson); 44 pushup(rt); 45 } 46 inline void Query(int val,int l,int r,int rt,int &p){ 47 if(tr[rt].lsum >= val){ 48 p = tr[rt].l; 49 return; 50 } 51 pushdown(rt); 52 int m = (l + r) >> 1; 53 if(tr[rt << 1].mi >= val) Query(val, lson, p); 54 else if(tr[rt].res >= val) p = tr[rt].p; 55 else if(tr[rt << 1 | 1].mi >= val) Query(val, rson, p); 56 //pushup(rt); 57 } 58 inline void update(int L,int R,int c,int l,int r,int rt){ 59 if(l >= L && r <= R){ 60 tr[rt].lsum =tr[rt].res = tr[rt].rsum = tr[rt].sum = tr[rt].mi = (tr[rt].r - tr[rt].l + 1) * c; 61 tr[rt].p = tr[rt].l; 62 col[rt] = c; 63 return; 64 } 65 pushdown(rt); 66 int m = (l + r) >> 1; 67 if(L <= m) update(L,R,c,lson); 68 if(R > m) update(L,R,c,rson); 69 pushup(rt); 70 } 71 int main(){ 72 std::ios::sync_with_stdio(false); 73 int opt, x, d, pos; 74 cin >> n >> Q; 75 build(1, n, 1); 76 for(int i = 1; i <= Q; i ++){ 77 cin >> opt; 78 if(opt == 1){ 79 cin >> d; 80 if(tr[1].mi < d){ 81 cout << 0 << endl; 82 } 83 else{ 84 Query(d,1,n,1,pos); 85 cout << pos << endl; 86 update(pos, pos + d - 1, 0, 1, n, 1); 87 } 88 } 89 else{ 90 cin >> x >> d; 91 update(x, x + d - 1, 1, 1, n, 1); 92 } 93 }return 0; 94 }
