Codeforces Round #222 (Div. 1) B - Preparing for the Contest

先二分，输入排序，然后对于确定的天数，贪心判断是否可行。

#include <iostream>#include <queue>#include <stack>#include <map>#include <set>#include <bitset>#include <cstdio>#include <algorithm>#include <cstring>#include <climits>#include <cstdlib>#include <cmath>#include <time.h>#define maxn 500005#define maxm 300005#define eps 1e-3#define mod 9999677#define INF 0x3f3f3f3f#define PI (acos(-1.0))#define lowbit(x) (x&(-x))#define mp make_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid #define rson o<<1 | 1, mid+1, R#define pii pair<int, int>#pragma comment(linker, "/STACK:16777216")typedef long long LL;typedef unsigned long long ULL;using namespace std;LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}//headstruct node{int a, b, c, id;}aa[maxn], a[maxn];set<pair<int, int> > s;int ans[maxn];int n, m, price;void read(){scanf("%d%d%d", &n, &m, &price);for(int i = 0; i < m; i++) scanf("%d", &a[i].a), a[i].id = i;for(int i = 0; i < n; i++) scanf("%d", &aa[i].b);for(int i = 0; i < n; i++) scanf("%d", &aa[i].c), aa[i].id = i;}int cmp1(node a, node b){return a.a > b.a;}int cmp2(node a, node b){return a.b > b.b;}bool check(int x){int ok = 1, cnt = price;s.clear();for(int i = 0, j = 0; i < m && ok; i += x) {while(j < n && aa[j].b >= a[i].a) s.insert(mp(aa[j].c, aa[j].id)), j++;if(!s.empty() && cnt >= s.begin()->first) cnt -= s.begin()->first, s.erase(s.begin());else ok = 0;}return ok;}void work(){sort(a, a+m, cmp1);sort(aa, aa+n, cmp2);int bot = 1, top = m, mid, res = -1;while(top >= bot) {int mid = (top + bot) >> 1;if(check(mid)) res = mid, top = mid-1;else bot = mid+1;}if(res == -1) printf("NO\n");else {printf("YES\n");int cnt = price;s.clear();for(int i = 0, j = 0; i < m; i += res) {while(j < n && aa[j].b >= a[i].a) s.insert(mp(aa[j].c, aa[j].id)), j++;int t = s.begin()->second;for(int k = i; k < m && k < i + res; k++) ans[a[k].id] = t;cnt -= s.begin()->first, s.erase(s.begin());}for(int i = 0; i < m; i++) printf("%d%c", ++ans[i], i == m-1 ? '\n' : ' ');}}int main(){read();work();return 0;}