BZOJ1222 [HNOI2001]产品加工

题解

怎么看都不像是个背包，直到我看了题解→_→， 第一次碰到这么奇怪的背包= = 

定一个滚动数组$F_i$， $i$表示机器$a$用了$i$的时间， $F_i$表示机器$b$用了$F_i$的时间， 然后就直接可以进行滚动转移了。。。

真是开眼界啊QuQ

代码

#include<cstring>
#include<algorithm>
#include<cstdio>
#define rd read()
#define rep(i,a,b) for(register int i = (a); i <= (b); ++i)
#define per(i,a,b) for(register int i = (a); i >= (b); --i)
using namespace std;

const int N = 8e3, inf = ~0U >> 3;

int n, m, f[N << 2];
int a[N], b[N], ab[N];

inline int read() {
    int X = 0, p = 1; char c = getchar();
    for(; c > '9' || c < '0'; c = getchar()) if(c == '-') p = -1;
    for(; c >= '0' && c <= '9'; c = getchar()) X = X * 10 + c - '0';
    return X * p;
}

inline int cmax(int A, int B) {
    return A > B ? A : B;
}

inline int cmin(int A, int B) {
    return A > B ? B : A;
}

int main()
{
    n = rd;
    rep(i, 1, n) {
        a[i] = rd;
        b[i] = rd;
        ab[i] = rd;
        m += cmax(a[i], ab[i]);
    }
    f[0] = 0;
    rep(i, 1, n) per(j, m, 0) {
        int tmp = inf;
        if(a[i] && j - a[i] >= 0) tmp = cmin(tmp, f[j - a[i]]);
        if(ab[i] && j - ab[i] >= 0) tmp = cmin(tmp, f[j - ab[i]] + ab[i]);
        if(b[i]) tmp = cmin(tmp, f[j] + b[i]);
        f[j] = tmp;
    }
    int ans = inf;
    rep(i ,0, m) ans = cmin(ans, cmax(i, f[i]));
    printf("%d
", ans);
}