When the monkey professor leaves his class for a short time, all the monkeys go bananas. N monkeys are lined up sitting side by side on their chairs. They each have the same joke book. Before the professor returns, M jokes were heard.
Each of the M jokes are said in the order given and have the following properties:
xi - position of the monkey who said it.
li – index of the joke in the book.
ki – volume the monkey says that joke.
When the monkey at position xi says the joke li, all monkeys at a distance less than or equal to ki from that monkey (including the monkey who said the joke) will fall off their chairs in laughter if they have never heard the joke li before.
If the joke li has been heard anytime during the past before, and the monkey hears it again, then he will sit back up in his chair.
A monkey can fall off his chair more than once (every time he hears a new joke), and if he is already on the ground and hears a new joke, he will stay on the ground.
Can you figure out how many monkeys will be in their seats by the time the professor comes back?
The first line of input is T – the number of test cases.
The first line of each test case is N, M (1 ≤ N ≤ 105) (1 ≤ M ≤ 105) – the number of monkeys in the class, and the number of jokes said before the professor returns.
The next M lines contain the description of each joke: xi, li, ki(1 ≤ xi ≤ N) (1 ≤ li ≤ 105) (0 ≤ ki ≤ N).
For each test case, output on a line a single integer - the number of monkeys in their seats after all jokes have been said.
1 10 7 3 11 0 3 11 2 5 12 1 8 13 2 7 11 2 10 12 1 9 12 0
3
【题解】
考虑 每一个人 站着还是坐下 取决于 最后一个他听到的笑话这个可以线段树nlogn求出然后我们依次考虑每一个笑话看他覆盖的区间此笑话决定的人 那个人被覆盖0或>1次则坐下 否则站起来也可以线段树综上 复杂度nlogn
（这是我跟某dalao的聊天记录直接放这里了）
网上有用扫描线做的，不是很明白。。。
待我理解理解
1 #include2 #include 3 #include 4 #include 5 #include 6 #include 7 #define min(a, b) ((a) < (b) ? (a) : (b)) 8 #define max(a, b) ((a) > (b) ? (a) : (b)) 9 10 inline void swap(long long &x, long long &y) 11 { 12 long long tmp = x;x = y;y = tmp; 13 } 14 15 inline void read(long long &x) 16 { 17 x = 0;char ch = getchar(), c = ch; 18 while(ch < '0' || ch > '9')c = ch, ch = getchar(); 19 while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar(); 20 if(c == '-')x = -x; 21 } 22 23 const long long INF = 0x3f3f3f3f; 24 const long long MAXN = 1000000 + 10; 25 26 struct Node 27 { 28 long long x, l, k, t; 29 }node[MAXN]; 30 31 long long n, m, l[MAXN], r[MAXN], tot, cnt[MAXN]; 32 33 34 //第一颗线段树，用来维护最晚时间颜色coloe 35 long long color[MAXN], num[MAXN]; 36 37 void pushup1(long long o, long long l, long long r) 38 { 39 long long mid = (l + r) >> 1; 40 if(!color[o << 1]) color[o << 1] = color[o]; 41 if(!color[o << 1 | 1]) color[o << 1 | 1] = color[o]; 42 } 43 44 void modify1(long long ll, long long rr,long long col, long long o = 1, long long l = 1, long long r = n) 45 { 46 pushup1(o, l, r); 47 if(color[o]) return; 48 if(ll <= l && rr >= r) 49 { 50 if(!color[o])color[o] = col; 51 pushup1(o, l, r); 52 return; 53 } 54 long long mid = (l + r) >> 1; 55 if(mid >= ll)modify1(ll, rr, col, o << 1, l, mid); 56 if(mid < rr) modify1(ll, rr, col, o << 1 | 1, mid + 1, r); 57 } 58 59 void build1(long long o = 1, long long l = 1, long long r = n) 60 { 61 if(l == r) 62 { 63 num[l] = color[o]; 64 return; 65 } 66 pushup1(o, l, r); 67 long long mid = (l + r) >> 1; 68 build1(o << 1, l, mid); 69 build1(o << 1 | 1, mid + 1, r); 70 } 71 72 //第二颗线段树，维护特定颜色作用于某个点多少次 73 long long data[MAXN], lazy[MAXN]; 74 75 void pushup2(long long o, long long l, long long r) 76 { 77 if(lazy[o]) 78 { 79 long long mid = (l + r) >> 1; 80 lazy[o << 1] += lazy[o]; 81 lazy[o << 1 | 1] += lazy[o]; 82 data[o << 1] += lazy[o] * (mid - l + 1); 83 data[o << 1 | 1] += lazy[o] * (r - mid); 84 lazy[o] = 0; 85 } 86 } 87 88 void modify2(long long ll, long long rr, long long x, long long o = 1, long long l = 1, long long r = n) 89 { 90 pushup2(o, l, r); 91 if(ll <= l && rr >= r) 92 { 93 lazy[o] += x; 94 data[o] += (r - l + 1) * x; 95 return; 96 } 97 long long mid = (l + r) >> 1; 98 if(mid >= ll)modify2(ll, rr, x, o << 1, l, mid); 99 if(mid < rr) modify2(ll, rr, x, o << 1 | 1, mid + 1, r);100 data[o] = data[o << 1] + data[o << 1 | 1];101 }102 103 long long ask2(long long p, long long o = 1, long long l = 1, long long r = n)104 {105 if(l == r && l == p)106 return data[o];107 pushup2(o, l, r);108 long long mid = (l + r) >> 1;109 if(p <= mid)return ask2(p, o << 1, l, mid);110 else return ask2(p, o << 1 | 1, mid + 1, r);111 }112 113 long long cmppp(Node a, Node b)114 {115 return a.l < b.l; 116 }117 118 long long cmp(Node a, Node b)119 {120 return a.t > b.t; 121 }122 123 long long cmpp(long long a, long long b)124 {125 return num[a] < num[b];126 }127 128 long long ans, t;129 130 int main()131 {132 // freopen("data.txt", "r", stdin);133 /*long long n, m,cnt[MAXN];134 */135 read(t);136 for(;t;-- t)137 {138 ans = 0;tot = 0;139 memset(color, 0, sizeof(color));140 memset(num, 0, sizeof(num));141 memset(data, 0, sizeof(data));142 memset(lazy, 0, sizeof(lazy));143 memset(l, 0, sizeof(l));144 memset(r, 0, sizeof(r));145 memset(node, 0, sizeof(node));146 memset(cnt, 0, sizeof(cnt));147 read(n), read(m);148 for(register long long i = 1;i <= m;++ i)149 read(node[i].x), read(node[i].l), read(node[i].k), node[i].t = i;150 for(register long long i = 1;i <= n;++ i) cnt[i] = i;151 for(register long long i = m;i >= 1;-- i)152 modify1(node[i].x - node[i].k, min(n, node[i].x + node[i].k), node[i].l);153 build1();154 std::sort(node + 1, node + 1 + m, cmppp);155 std::sort(cnt + 1, cnt + 1 + n, cmpp);156 long long now = node[1].l;tot = 1;l[1] = 1;157 for(register long long i = 2;i <= m;++ i)158 if(now != node[i].l)r[tot] = i - 1, ++ tot, now = node[i].l, l[tot] = i;159 r[tot] = m;160 for(register long long i = 1, p = 1;i <= tot;++ i)161 {162 for(register long long j = l[i];j <= r[i];++ j)163 modify2(node[j].x - node[j].k, min(n, node[j].x + node[j].k), 1);164 long long tmp = 0, flag = p;165 while(num[cnt[p]] == node[l[i]].l)166 {167 long long tmp = ask2(cnt[p]);168 if(tmp == 0 || tmp > 1) ++ ans;169 ++ p;170 }171 for(register long long j = l[i];j <= r[i];++ j)172 modify2(node[j].x - node[j].k, min(n, node[j].x + node[j].k), -1);173 }174 printf("%I64d\n", ans);175 }176 return 0;177 }