线段树区间合并，这题写起来还是有点麻烦的，又需要lazy标记，又需要区间合并，不过还是1A，多写数据结构的题目果真能锻炼代码能力。
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 using namespace std; 5 6 const int N = 50001; 7 8 struct Node 9 { 10 int l, r, lazy; 11 int len, llen, rlen; 12 } node[N << 2]; 13 14 void build( int i, int l, int r ) 15 { 16 node[i].l = l, node[i].r = r; 17 node[i].lazy = -1; 18 node[i].len = node[i].llen = node[i].rlen = r - l + 1; 19 if ( l == r ) return ; 20 int mid = ( l + r ) >> 1; 21 build( i << 1, l, mid ); 22 build( i << 1 | 1, mid + 1, r ); 23 } 24 25 void pushup( int i ) 26 { 27 int lc = i << 1, rc = lc | 1; 28 node[i].llen = node[lc].llen; 29 if ( node[lc].len == node[lc].r - node[lc].l + 1 ) 30 { 31 node[i].llen += node[rc].llen; 32 } 33 node[i].rlen = node[rc].rlen; 34 if ( node[rc].len == node[rc].r - node[rc].l + 1 ) 35 { 36 node[i].rlen += node[lc].rlen; 37 } 38 node[i].len = max( node[lc].len, node[rc].len ); 39 node[i].len = max( node[i].len, node[lc].rlen + node[rc].llen ); 40 } 41 42 void pushdown( int i ) 43 { 44 if ( node[i].lazy == -1 ) return ; 45 int lc = i << 1, rc = lc | 1; 46 node[lc].lazy = node[rc].lazy = node[i].lazy; 47 if ( node[i].lazy == 0 ) 48 { 49 node[lc].len = node[lc].llen = node[lc].rlen = node[lc].r - node[lc].l + 1; 50 node[rc].len = node[rc].llen = node[rc].rlen = node[rc].r - node[rc].l + 1; 51 } 52 else 53 { 54 node[lc].len = node[lc].llen = node[lc].rlen = 0; 55 node[rc].len = node[rc].llen = node[rc].rlen = 0; 56 } 57 node[i].lazy = -1; 58 } 59 60 void update( int i, int l, int r, int c ) 61 { 62 if ( node[i].l == l && node[i].r == r ) 63 { 64 node[i].lazy = c; 65 if ( c == 0 ) 66 { 67 node[i].len = node[i].llen = node[i].rlen = r - l + 1; 68 } 69 else 70 { 71 node[i].len = node[i].llen = node[i].rlen = 0; 72 } 73 return ; 74 } 75 pushdown(i); 76 int mid = ( node[i].l + node[i].r ) >> 1; 77 if ( r <= mid ) 78 { 79 update( i << 1, l, r, c ); 80 } 81 else if ( l > mid ) 82 { 83 update( i << 1 | 1, l, r, c ); 84 } 85 else 86 { 87 update( i << 1, l, mid, c ); 88 update( i << 1 | 1, mid + 1, r, c ); 89 } 90 pushup(i); 91 } 92 93 int query( int i, int d ) 94 { 95 if ( node[i].l == node[i].r ) return node[i].l; 96 pushdown(i); 97 int lc = i << 1, rc = lc | 1; 98 if ( node[lc].len >= d ) 99 { 100 return query( lc, d ); 101 } 102 else if ( node[lc].rlen + node[rc].llen >= d ) 103 { 104 return node[lc].r - node[lc].rlen + 1; 105 } 106 else 107 { 108 return query( rc, d ); 109 } 110 } 111 112 int main () 113 { 114 int n, m; 115 while ( scanf("%d%d", &n, &m) != EOF ) 116 { 117 build( 1, 1, n ); 118 int op, x, y; 119 while ( m-- ) 120 { 121 scanf("%d", &op); 122 if ( op == 1 ) 123 { 124 int d; 125 scanf("%d", &d); 126 if ( node[1].len < d ) 127 { 128 printf("0 "); 129 continue; 130 } 131 x = query( 1, d ); 132 printf("%d ", x); 133 y = x + d - 1; 134 update( 1, x, y, 1 ); 135 } 136 else 137 { 138 scanf("%d%d", &x, &y); 139 y += x - 1; 140 update( 1, x, y, 0 ); 141 } 142 } 143 } 144 return 0; 145 }