cf1555 E. Boring Segments

题意：

给你n个线段，最大点是m，每一个线段有一个权值w，你能选择线段来覆盖1-m这个区间的，选择的代价为最大权值和最小权值的差。问你最小的的代价是多少。

题解：

尺取+线段树
我们尺取的选择线段，然后用线段树来判断此时区间是否被完全覆盖，如何判断呢？我们可以认为一开始整个线段树都是0，每加入一个线段，这个区间的值+1，当tr[1].sum!=0时，即所有点都被覆盖。
注意：本题中的覆盖是覆盖所有边，比如线段[1,5]和线段[6,10]并没有将[1,10]这个范围覆盖，应该是[1,5]和[5,10]才算覆盖，所以我们可以将m个点转换成m-1个边，覆盖这m-1个边，每个线段的右端点也要-1

代码：

// Problem: E. Boring Segments
// Contest: Codeforces - Educational Codeforces Round 112 (Rated for Div. 2)
// URL: https://codeforces.com/contest/1555/problem/E
// Memory Limit: 256 MB
// Time Limit: 3000 ms
// Data:2021-08-18 13:52:58
// By Jozky#include <bits/stdc++.h>
#include <unordered_map>
#define debug(a, b) printf("%s = %d\n", a, b);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
clock_t startTime, endTime;
//Fe~Jozky
const ll INF_ll= 1e18;
const int INF_int= 0x3f3f3f3f;
void read(){};
template <typename _Tp, typename... _Tps> void read(_Tp& x, _Tps&... Ar)
{x= 0;char c= getchar();bool flag= 0;while (c < '0' || c > '9')flag|= (c == '-'), c= getchar();while (c >= '0' && c <= '9')x= (x << 3) + (x << 1) + (c ^ 48), c= getchar();if (flag)x= -x;read(Ar...);
}
template <typename T> inline void write(T x)
{if (x < 0) {x= ~(x - 1);putchar('-');}if (x > 9)write(x / 10);putchar(x % 10 + '0');
}
void rd_test()
{
#ifdef LOCALstartTime= clock();freopen("in.txt", "r", stdin);
#endif
}
void Time_test()
{
#ifdef LOCALendTime= clock();printf("\nRun Time:%lfs\n", (double)(endTime - startTime) / CLOCKS_PER_SEC);
#endif
}
const int maxn= 1e6 + 9;
struct node
{int l, r, w;
} a[maxn];
bool cmp(node a, node b)
{return a.w < b.w;
}
struct tree
{int l, r;int minn;int lazy;
} tr[maxn << 2];
void solve(int rt, int val)
{tr[rt].lazy+= val;tr[rt].minn+= val;
}
void pushdown(int rt)
{solve(rt << 1, tr[rt].lazy);solve(rt << 1 | 1, tr[rt].lazy);tr[rt].lazy= 0;
}
void pushup(int rt)
{tr[rt].minn= min(tr[rt << 1].minn, tr[rt << 1 | 1].minn);
}
void update(int rt, int l, int r, int val)
{if (tr[rt].l > r || tr[rt].r < l)return;if (tr[rt].l >= l && tr[rt].r <= r) {solve(rt, val);return;}if (tr[rt].lazy)pushdown(rt);update(rt << 1, l, r, val);update(rt << 1 | 1, l, r, val);pushup(rt);
}
void build(int rt, int l, int r)
{//cout<<"rt="<<rt<<endl;tr[rt].l= l;tr[rt].r= r;tr[rt].lazy= 0;if (l == r) {tr[rt].minn= 0;return;}int mid= (l + r) >> 1;build(rt << 1, l, mid);build(rt << 1 | 1, mid + 1, r);pushup(rt);
}
int main()
{//rd_test();int n, m;read(n, m);for (int i= 1; i <= n; i++) {int l, r, w;read(l, r, w);a[i]= {l, --r, w};}m--;sort(a + 1, a + 1 + n, cmp);build(1, 1, m);int minn= INF_int;for (int i= 1, j= 0; i <= n; i++) {while (j <= n && tr[1].minn <= 0) {j++;update(1, a[j].l, a[j].r, 1);}if (j >= i && tr[1].minn)minn= min(minn, a[j].w - a[i].w);update(1, a[i].l, a[i].r, -1);}printf("%d\n", minn);//Time_test();
}