cf1555 E. Boring Segments

題意：

給你n個(gè)線段，最大點(diǎn)是m，每一個(gè)線段有一個(gè)權(quán)值w，你能選擇線段來(lái)覆蓋1-m這個(gè)區(qū)間的，選擇的代價(jià)為最大權(quán)值和最小權(quán)值的差。問(wèn)你最小的的代價(jià)是多少。

題解：

尺取+線段樹(shù)
我們尺取的選擇線段，然后用線段樹(shù)來(lái)判斷此時(shí)區(qū)間是否被完全覆蓋，如何判斷呢？我們可以認(rèn)為一開(kāi)始整個(gè)線段樹(shù)都是0，每加入一個(gè)線段，這個(gè)區(qū)間的值+1，當(dāng)tr[1].sum!=0時(shí)，即所有點(diǎn)都被覆蓋。
注意：本題中的覆蓋是覆蓋所有邊，比如線段[1,5]和線段[6,10]并沒(méi)有將[1,10]這個(gè)范圍覆蓋，應(yīng)該是[1,5]和[5,10]才算覆蓋，所以我們可以將m個(gè)點(diǎn)轉(zhuǎn)換成m-1個(gè)邊，覆蓋這m-1個(gè)邊，每個(gè)線段的右端點(diǎn)也要-1

代碼：

// 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(); }

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