BZOJ1021 SHOI2008循环的债务

dp模拟即可。

d[i][j][k]表示使用前i种面值，1号手里钱为j，2号手里钱为k时最少操作数

使用滚动数组压缩空间

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int n = ;
const int M = ;
const int val[n] = {, , , , , };
int x1, x2, x3;
int now, tot;
int suma, sumb, dis;
int sum[], Cnt[n];
int d[][M][M], cnt[][n];
int gcd(int a,int b)
{
    return b==?a:gcd(b,a%b);
}
inline void update(int &x, int y){
    if (x == -) x = y;
    else x = min(x, y);
}

inline int calc(int i,int a,int b){
    return (abs(a - cnt[][i]) + abs(b - cnt[][i]) + abs(Cnt[i] - a - b - cnt[][i])) / ;
}

int main(){
    scanf("%d%d%d", &x1, &x2, &x3);
    for (int i = ; i < ; ++i){
        sum[i] = ;
        for (int j = n - ; j >= ; --j){
            scanf("%d", cnt[i] + j);
            Cnt[j] += cnt[i][j];
            sum[i] += cnt[i][j] * val[j];
        }
        tot += sum[i];
    }
    int ea = sum[], eb = sum[], ec = sum[];
    ea += x3 - x1, eb += x1 - x2, ec += x2 - x3;
    if (ea <  || eb <  || ec <  || ea + eb + ec != tot)
    {printf("impossible\n");return ;}
    memset(d[], -, sizeof(d[]));
    d[][sum[]][sum[]] = ;
    for (int i = ; i < n; ++i)
    {
        now = i & ;
        int g=val[i];
        for(int j=i+;j<n;++j)
        g=gcd(g,val[j]);
        memset(d[now ^ ], -, sizeof(d[now ^ ]));
        for (int j = ; j <= tot; ++j)
        {
            for (int k = ; j + k <= tot; ++k)
            {
                if (d[now][j][k] >= )
                {
                    update(d[now ^ ][j][k], d[now][j][k]);
                    for (int a = ; a <= Cnt[i]; ++a)
                    {
                        for (int b = ; a + b <= Cnt[i]; ++b)
                        {
                            suma = j + val[i] * (a - cnt[][i]);
                            sumb = k + val[i] * (b - cnt[][i]);
                            if (suma >=  && sumb >=  && suma + sumb <= tot)
                            {
                                dis = calc(i,a,b);
                                update(d[now ^ ][suma][sumb], d[now][j][k] + dis);
                            }
                        }
                    }
                }
            }
        }
    }
    if(d[n&][ea][eb]<) puts("impossible");
    else printf("%d\n", d[n & ][ea][eb]);
    return ;
}