题目链接
题解
挺神的贪心
把第次买的蔬菜拆出来，记下每种蔬菜到期的日期，填第一单位蔬菜比其他的要晚
按价格排序后，贪心的往前面可以填的位置填就可以了。找可以填的位置用并查集维护一下。这样就求出了最大天数的答案。
对于询问的答案，从最后一天往前推，把最便宜的那些丢掉就好了。
代码
#include<cstdio>
#include<cstring>
#include<algorithm>
#define gc getchar
#define pc putchar
#define int long long
inline int read() {
int x = 0,f = 1;
char c = getchar();
while(c < '0' || c > '9') c = gc();
while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = gc();
return x * f;
}
void print(int x) {
if(x < 0) {
pc('-'); x = -x;
}
if(x >= 10)print(x / 10);
pc(x % 10 + '0');
}
const int maxn = 200007;
int n,m,k;
struct node {
int a,d,c,x;
node(int a = 0,int c = 0,int x = 0,int d = 0) : a(a), c(c),x(x),d(d) {};
bool operator < (const node &k) const {
return a > k.a;
}
} a[maxn];
int tot = 0;
int q[maxn],fa[maxn];
int find(int x ) {
if(fa[x] != x) fa[x] = find(fa[x]);
return fa[x];
}
void unionn(int x,int y) {
int fx = find(x),fy = find(y);
if(fx == fy) return ;
fa[fy] = fx;
}
int cnt = 0;
int g[maxn],d[maxn],ga[maxn];
int sum = 0;
void calc(int idx,int c,int a) {
sum += 1ll * c * a;
g[cnt] += c;ga[cnt] = a;
d[idx] += c;
if(d[idx] == m) unionn(idx - 1,idx);
}
int L[maxn],upd[maxn];
int ans[maxn];
main() {
//freopen("1.in","r",stdin);
n = read(),m = read(),k = read();
for(int aa,s,c,x,i = 1;i <= n;++ i) {
aa = read(),s = read(),c = read(),x = read();
a[++ tot] = node(aa + s,1,0,x ? (c - 1) / x + 1 : maxn - 7);
if(-- c) a[++ tot] = node(aa,c,x,x ? (c - 1) / x + 1 : maxn - 7);
}
int maxd = 0;
for(int i = 1;i <= k;++ i) maxd = std::max(maxd,q[i] = read());
std::sort(a + 1,a + tot + 1);
for(int i = 1;i <= maxd;++ i) fa[i] = i;
for(int i = 1;i <= tot;++ i) {
cnt ++;
int idx = find(std::min(a[i].d,maxd)); //a 价值 d GG日期 d 数量 x 每天GG数
int res = (idx - 1) * a[i].x ,now = a[i].c - res;
while(idx && now) {
int mn = std::min(m - d[idx],now);
calc(idx,mn,a[i].a);
now -= mn;
int p = idx;
idx = find(idx - 1);
p -= idx;
if(res) now += p * a[i].x , res -= p * a[i].x;
}
if(!find(maxd)) break ;
}
int R = 0;
for(int i = 1;i <= maxd;++ i) L[i] = m - d[i] + L[i - 1];
for(int i = maxd;i;-- i) upd[i] = std::max(R - L[i],0ll),R += d[i];
for(int i = maxd;i >= 1;-- i) {
ans[i] = sum;
int tmp = upd[i - 1] - upd[i];
while(tmp) {
int mn = std::min(tmp,g[cnt]);
sum -= 1ll * ga[cnt] * mn;
tmp -= mn;
g[cnt] -= mn;
if(!g[cnt]) -- cnt;
}
}
for(int i = 1;i <= k;++ i) print(ans[q[i]]),pc('
');// print(10);
return 0;
} // 1110001