题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4614
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较为麻烦的一道线段树题目 为了更方便处理可以把
查询左端点 查询右端点 区间查询 这三个查询分开写
最后复杂度是 $O(mlog(n))$的 $($下面的代码就是用的这个思路$)$
1 #include <cstdio> 2 #include <cstring> 3 #include <cmath> 4 #include <algorithm> 5 using namespace std; 6 const int N = 50010; 7 int sum[N << 2], flag[N << 2]; 8 int t, n, m; 9 void pushdown(int x, int tl, int tr) 10 { 11 if(!flag[x]) 12 return; 13 flag[x << 1] = flag[x << 1 | 1] = flag[x]; 14 if(flag[x] & 1) 15 sum[x << 1] = sum[x << 1 | 1] = 0; 16 else 17 { 18 int mid = (tl + tr) >> 1; 19 sum[x << 1] = mid - tl + 1; 20 sum[x << 1 | 1] = tr - mid; 21 } 22 flag[x] = 0; 23 } 24 void update(int x, int L, int R, int tl, int tr, int y) 25 { 26 if(L <= tl && tr <= R) 27 { 28 flag[x] = y; 29 if(y & 1) 30 sum[x] = 0; 31 else 32 sum[x] = tr - tl + 1; 33 return; 34 } 35 int mid = (tl + tr) >> 1; 36 pushdown(x, tl, tr); 37 if(L <= mid) 38 update(x << 1, L, R, tl, mid, y); 39 if(mid < R) 40 update(x << 1 | 1, L, R, mid + 1, tr, y); 41 sum[x] = sum[x << 1] + sum[x << 1 | 1]; 42 } 43 int queryl(int x, int L, int tl, int tr) 44 { 45 if(!sum[x]) 46 return -1; 47 if(tl == tr) 48 return tl; 49 int mid = (tl + tr) >> 1; 50 pushdown(x, tl, tr); 51 int re = -1; 52 if(L <= mid && sum[x << 1]) 53 re = queryl(x << 1, L, tl, mid); 54 if(re != -1) 55 return re; 56 return queryl(x << 1 | 1, L, mid + 1, tr); 57 } 58 int queryr(int x, int L, int tl, int tr, int &num) 59 { 60 if(tl == tr) 61 { 62 num -= sum[x]; 63 return tl; 64 } 65 int mid = (tl + tr) >> 1; 66 pushdown(x, tl, tr); 67 if(L <= tl) 68 { 69 if(sum[x << 1] >= num || !sum[x << 1 | 1]) 70 return queryr(x << 1, L, tl, mid, num); 71 num -= sum[x << 1]; 72 return queryr(x << 1 | 1, L, mid + 1, tr, num); 73 } 74 if(L <= mid) 75 { 76 int re = queryr(x << 1, L, tl, mid, num); 77 if(!num || !sum[x << 1 | 1]) 78 return re; 79 } 80 return queryr(x << 1 | 1, L, mid + 1, tr, num); 81 } 82 int query(int x, int L, int R, int tl, int tr) 83 { 84 if(L <= tl && tr <= R) 85 return tr - tl + 1 - sum[x]; 86 int mid = (tl + tr) >> 1, re = 0; 87 pushdown(x, tl, tr); 88 if(L <= mid) 89 re += query(x << 1, L, R, tl, mid); 90 if(mid < R) 91 re += query(x << 1 | 1, L, R, mid + 1, tr); 92 return re; 93 } 94 int main() 95 { 96 scanf("%d", &t); 97 while(t--) 98 { 99 scanf("%d%d", &n, &m); 100 memset(sum, 0, sizeof sum); 101 memset(flag, 0, sizeof flag); 102 update(1, 0, n - 1, 0, n - 1, 2); 103 int x, y, z; 104 while(m--) 105 { 106 scanf("%d%d%d", &x, &y, &z); 107 if(x == 1) 108 { 109 int L = queryl(1, y, 0, n - 1); 110 if(L == -1) 111 { 112 puts("Can not put any one."); 113 continue; 114 } 115 int R = queryr(1, L, 0, n - 1, z); 116 printf("%d %d ", L, R); 117 update(1, L, R, 0, n - 1, 1); 118 } 119 else 120 { 121 printf("%d ", query(1, y, z, 0, n - 1)); 122 update(1, y, z, 0, n - 1, 2); 123 } 124 } 125 puts(""); 126 } 127 return 0; 128 }
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不过如果不想写这么多判断的话也可以多写个二分 这样就只要区间查询
然而复杂度就会变成$O(mlog^2(n))$