BZOJ1021 [SHOI2008]Debt 循环的债务

貌似去年暑假就听过这道题。。。那时候还YY了个什么平面三条轴，夹角Π/3之类的。。。

正解嘛。。。当然是DP

令f[i][j][k]表示到了第i种面值，第一个人还有j元钱，第二个人还有k元钱的最少交换张数。

于是就是个背包问题的说，然后因为dp方程太复杂了，请参考程序吧。。。

（还有个非常厉害的剪枝我的程序没加，因为太懒了≥v≤。。。。。。誒！不要打我啊~~~都过了嘛好不好QAQ）

/**************************************************************
    Problem: 1021
    User: rausen
    Language: C++
    Result: Accepted
    Time:516 ms
    Memory:8696 kb
****************************************************************/
 
#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 i, j, k, a, b, suma, sumb, dis;
int sum[3], Cnt[n];
int d[2][M][M], cnt[3][n];
 
inline void update(int &x, int y){
    if (x == -1) x = y;
    else x = min(x, y);
}
 
inline int calc(){
    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 (i = 0; i < 3; ++i){
        sum[i] = 0;
        for (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];
    }
     
    memset(d[0], -1, sizeof(d[0]));
    d[0][sum[0]][sum[1]] = 0;
    for (i = 0; i < n; ++i){
        now = i & 1;
        memset(d[now ^ 1], -1, sizeof(d[now ^ 1]));
        for (j = 0; j <= tot; ++j){
            for (k = 0; j + k <= tot; ++k){
                if (d[now][j][k] >= 0){
                    update(d[now ^ 1][j][k], d[now][j][k]);
                    for (a = 0; a <= Cnt[i]; ++a){
                        for (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();
                                update(d[now ^ 1][suma][sumb], d[now][j][k] + dis);
                            }
                        }
                    }
                }
            }
        }
    }
    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 || d[n & 1][ea][eb] < 0)
        printf("impossible
");
    else printf("%d
", d[n & 1][ea][eb]);
    return 0;
}