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 = 6;
const int M = 1005;
const int val[n] = {1, 5, 10, 20, 50, 100};
int x1, x2, x3;
int now, tot;
int suma, sumb, dis;
int sum[3], Cnt[n];
int d[2][M][M], cnt[3][n];
int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}
inline void update(int &x, int y){
    if (x == -1) x = y;
    else x = min(x, y);
}

inline int calc(int i,int a,int b){
    return (abs(a - cnt[0][i]) + abs(b - cnt[1][i]) + abs(Cnt[i] - a - b - cnt[2][i])) / 2;
}

int main(){
    scanf("%d%d%d", &x1, &x2, &x3);
    for (int i = 0; i < 3; ++i){
        sum[i] = 0;
        for (int j = n - 1; j >= 0; --j){
            scanf("%d", cnt[i] + j);
            Cnt[j] += cnt[i][j];
            sum[i] += cnt[i][j] * val[j];
        }
        tot += sum[i];
    }
    int ea = sum[0], eb = sum[1], ec = sum[2];
    ea += x3 - x1, eb += x1 - x2, ec += x2 - x3;
    if (ea < 0 || eb < 0 || ec < 0 || ea + eb + ec != tot)
    {printf("impossible\n");return 0;}
    memset(d[0], -1, sizeof(d[0]));
    d[0][sum[0]][sum[1]] = 0;
    for (int i = 0; i < n; ++i)
    {
        now = i & 1;
        int g=val[i];
        for(int j=i+1;j<n;++j)
        g=gcd(g,val[j]);
        memset(d[now ^ 1], -1, sizeof(d[now ^ 1]));
        for (int j = 0; j <= tot; ++j)
        {
            for (int k = 0; j + k <= tot; ++k)
            {
                if (d[now][j][k] >= 0)
                {
                    update(d[now ^ 1][j][k], d[now][j][k]);
                    for (int a = 0; a <= Cnt[i]; ++a)
                    {
                        for (int b = 0; a + b <= Cnt[i]; ++b)
                        {
                            suma = j + val[i] * (a - cnt[0][i]);
                            sumb = k + val[i] * (b - cnt[1][i]);
                            if (suma >= 0 && sumb >= 0 && suma + sumb <= tot)
                            {
                                dis = calc(i,a,b);
                                update(d[now ^ 1][suma][sumb], d[now][j][k] + dis);
                            }
                        }
                    }
                }
            }
        }
    }
    if(d[n&1][ea][eb]<0) puts("impossible");
    else printf("%d\n", d[n & 1][ea][eb]);
    return 0;
}

原文地址：https://www.cnblogs.com/nbwzyzngyl/p/8318916.html