$des$
有 $n$ 个物品，第 $i$ 个物品的价格是 $v_i$ ，有两个人，每个人都喜欢 $n$ 个物品中的一些物品。
要求选出正好 $m$ 个物品，满足选出的物品中至少有 $k$ 个物品被第一个人喜欢，$k$ 个物品被第二个人喜欢。并求出最小的价格和。
$sol$
将所有物品分成 $4$ 类
1. $a \cap b$
2. $a - a \cap b$
3. $b - a \cap b$
4. $全集 - a \cup b$
枚举 $1$ 选了多少个
此时 $2, 3$ 选多少个是固定的了
这样的话只需维护剩余元素的前 $x$ 小值之和
对顶堆维护
大根堆维护前 $x$ 小
查询时就是大根堆的权值之和
删除元素标记
维护剩余元素的前 $x$ 小值之和也可以用权值线段树
#include <bits/stdc++.h> using namespace std; const int N = 2e5 + 10; #define gc getchar() inline int read() { int x = 0; char c = gc; while(c < '0' || c > '9') c = gc; while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc; return x; } #define Rep(i, a, b) for(int i = a; i <= b; i ++) #define LL long long #define Fin(a) freopen(a, "r", stdin) #define Fout(a) freopen(a, "w", stdout) #define E return #define End cout << "\n" map <int, int> Map; int W[N]; int n, m, k, a, b; bool usea[N], useb[N]; priority_queue <int, vector <int>, less <int> > Big; priority_queue <int, vector <int>, greater <int> > Small; int szab[N], jsab; int sza[N], jsa; int szb[N], jsb; int sznot[N], jsnot; LL sum1[N], sum2[N], sum3[N]; int Size; LL Sum; void Update() { while(Big.size() && Map[Big.top()]) Map[Big.top()] --, Big.pop(); while(Small.size() && Map[Small.top()]) Map[Small.top()] --, Small.pop(); } int Now_size; int ato, bto, abto; void Add(int x) { Update(); while(Size < Now_size && Small.size()) { int topsmall = Small.top(); Small.pop(); Size ++, Sum += topsmall, Big.push(topsmall); } if(Size == Now_size) { if(Big.size() == 0) Small.push(x); else if(x < Big.top()) { int topbig = Big.top(); Sum += (x - topbig); Big.pop(); Big.push(x); Small.push(topbig); } return ; } if(Size < Now_size) { Big.push(x); Size ++; Sum += x; return ; } Update(); int topbig = Big.top(); if(x < topbig) { Sum += (x - topbig); Big.pop(); Big.push(x); Small.push(topbig); } else { Small.push(x); } } bool Judge() { if(m < k) return 1; if(jsab + jsa < k || jsab + jsb < k) return 1; if(k * 2 - jsab > m) return 1; return 0; } void RE() { cout << "error" << "\n"; } void Bef_work() { Rep(i, 1, n) { if(usea[i] && useb[i]) szab[++ jsab] = W[i]; else if(usea[i]) sza[++ jsa] = W[i]; else if(useb[i]) szb[++ jsb] = W[i]; else sznot[++ jsnot] = W[i]; } sort(szab + 1, szab + jsab + 1); sort(sza + 1, sza + jsa + 1); sort(szb + 1, szb + jsb + 1); sort(sznot + 1, sznot + jsnot + 1); Rep(i, 1, jsab) sum1[i] = sum1[i - 1] + szab[i]; Rep(i, 1, jsa) sum2[i] = sum2[i - 1] + sza[i]; Rep(i, 1, jsb) sum3[i] = sum3[i - 1] + szb[i]; if(Judge()) {puts("-1"); exit(0) ;} Rep(i, 1, jsab) Add(szab[i]); Rep(i, k + 1, jsa) Add(sza[i]); Rep(i, k + 1, jsb) Add(szb[i]); Rep(i, 1, jsnot) Add(sznot[i]); ato = min(k, jsa); bto = min(k, jsb); abto = 0; } bool Judge2(int x) { if(k - x > jsa || k - x > jsb || k * 2 - x > m) return 0; return 1; } void Del(int x) { Update(); Map[x] ++; int topbig = Big.top(); if(x <= topbig) { Size --, Sum -= x; } Update(); } LL Ask(int x) { Update(); while(Size > x && Big.size()) { int topbig = Big.top(); Size --, Sum -= topbig; Big.pop(); Small.push(topbig); Update(); } while(Size < x && Small.size()) { int topsmall = Small.top(); Size ++, Sum += topsmall; Small.pop(); Big.push(topsmall); Update(); } return Sum; } LL Calc(int x) { LL ret = 0; while(abto <= x && abto) { Del(szab[abto]); abto ++; } ret += sum1[abto - 1]; while(ato > k - x && ato) { Add(sza[ato]); ato --; } ret += sum2[ato]; while(bto > k - x && bto) { Add(szb[bto]); bto --; } ret += sum3[bto]; ret += Ask(m - k * 2 + x); return ret; } int main() { n = read(), m = read(), k = read(); Rep(i, 1, n) W[i] = read(); a = read(); Rep(i, 1, a) W[0] = read(), usea[W[0]] = 1; b = read(); Rep(i, 1, b) W[0] = read(), useb[W[0]] = 1; Now_size = m; Bef_work(); abto = 1; LL Answer = 1e18; Rep(i, 0, jsab) { if(!Judge2(i)) continue; Now_size = m - k * 2 + i; Answer = min(Answer, Calc(i)); } cout << (Answer == 1e18 ? -1 : Answer); return 0; }
orz zbq
/* 枚举两个人都喜欢的物品选择了几个， 那么我们就知道了只有一个人喜欢的物品选择了几个 这两种显然贪心要最小的 然后剩下的都可以随便选 所以扔到数据结构里找前k小的和 枚举移动一位时， 线段树里的数据改动量是O1的即可 */ #include<cstdio> #include<algorithm> #include<cstring> #include<queue> #include<iostream> #define ll long long #define M 300010 #include<map> #define tmp tot using namespace std; int read() { int nm = 0, f = 1; char c = getchar(); for(; !isdigit(c); c = getchar()) if(c == '-') f = -1; for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0'; return nm * f; } ll n, m, k; int ver[M]; bool vis1[M], vis2[M]; ll st1[M], tp1, st2[M], tp2,st3[M], tp3,st4[M], tp4; ll ans = 1000000000000000010ll; ll allans = 0; ll note[M], biao[M]; map<ll, ll> mp; ll notev[M * 20]; ll t[M * 10], sz[M * 10]; #define lson l, mid, now << 1 #define rson mid + 1, r, now << 1 | 1 void insert(int l, int r, int now, int x) { if(l > x || r < x) return; if(l == r) { t[now] += biao[x]; sz[now]++; notev[now] = biao[x]; return; } int mid = (l + r) >> 1; insert(lson, x), insert(rson, x); t[now] = t[now << 1] + t[now << 1 | 1]; sz[now] = sz[now << 1] + sz[now << 1 | 1]; } void del(int l, int r, int now, int x) { if(l > x || r < x) return; if(l == r) { t[now] -= biao[x]; sz[now]--; return; } int mid = (l + r) >> 1; del(lson, x), del(rson, x); t[now] = t[now << 1] + t[now << 1 | 1]; sz[now] = sz[now << 1] + sz[now << 1 | 1]; } ll query(int k) { int now = 1; ll tmp = 0; while(k) { if(sz[now] == 0) break; if(sz[now] && !sz[now << 1] && !sz[now << 1 | 1])tmp += notev[now] * k, k = 0 /*-= sz[now]*/, now = now << 1 | 1; else if(sz[now << 1] <= k) tmp += t[now << 1], k-= sz[now << 1], now = now << 1 | 1; else now = now << 1; } return tmp; } int main() { n = read(), m = read(), k = read(); for(int i = 1; i <= n; i++) { note[i] = ver[i] = read(); } int q = read(); for(int i = 1; i <= q; i++) { int op = read(); vis1[op] = true; } q = read(); for(int i = 1; i <= q; i++) { int op = read(); vis2[op] = true; } for(int i = 1; i <= n; i++) { if(vis1[i] && vis2[i]) st1[++tp1] = ver[i]; else if(vis1[i]) st2[++tp2] = ver[i]; else if(vis2[i]) st3[++tp3] = ver[i]; else st4[++tp4] = ver[i]; } sort(st1 + 1, st1 + tp1 + 1); sort(st2 + 1, st2 + tp2 + 1); sort(st3 + 1, st3 + tp3 + 1); sort(st4 + 1, st4 + tp4 + 1); sort(note + 1, note + n + 1); note[0] = -1; int tot = 0; for(int i = 1; i <= n; i++) { if(note[i] == note[i - 1]) continue; tot++; biao[tot] = note[i]; mp[note[i]] = tot; } /* 前面是数据预处理， 离散化部分 */ int rx = min(k, tp1); int lx = max(0ll, max(k - tp2, k - tp3)); /* 确定m上下界， 不能弄出不够的情况 */ if(min(tp2, tp3) + tp1 < k || rx < lx) { printf("-1"); return 0; } for(int i = 1; i <= tp4; i++) insert(1, tmp, 1, mp[st4[i]]); for(int i = lx + 1; i <= tp1; i++) insert(1, tmp, 1, mp[st1[i]]); for(int i = k - lx + 1; i <= tp2; i++) insert(1, tmp, 1, mp[st2[i]]); for(int i = k - lx + 1; i <= tp3; i++) insert(1, tmp, 1, mp[st3[i]]); /* 将所有随便选的扔到线段树里 */ for(int i = 1; i <= lx; i++) allans += st1[i]; for(int i = 1; i <= k - lx; i++) allans += st2[i] + st3[i]; /* 统计当前答案 */ int tppp = k - lx; // 这个变量是一个人喜欢的选了几个 int zz = m - 2 * (k - lx) - lx; if(zz >= 0) ans = min(ans, allans + query(zz));//当前可行就统计答案 for(int i = lx + 1; i <= rx; i++) { // i是两个人喜欢的 del(1, tmp, 1, mp[st1[i]]); ;// 强制要选 所以从线段树中删除 allans += st1[i]; if(tppp == 0) break;//不合法 insert(1, tmp, 1, mp[st2[tppp]]), allans -= st2[tppp]; insert(1, tmp, 1, mp[st3[tppp]]), allans -= st3[tppp]; /* 将已经可以随便选的 一个人喜欢的加入线段树 */ tppp--; int op = m - 2 * tppp - i; //从线段树中要拿多少 if(op < 0) continue; if(op > sz[1]) continue; ans = min(ans, allans + query(op)); } if(ans == 1000000000000000010ll) ans = -1; cout << ans << "\n"; return 0; }