bzoj4946: [Noi2017]蔬菜 神烦贪心

题目链接

bzoj4946: [Noi2017]蔬菜

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('\n');// print(10);
    return 0;
} // 1110001