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）

 1 /**************************************************************
 2     Problem: 1021
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:516 ms
 7     Memory:8696 kb
 8 ****************************************************************/
 9
10 #include <cstdio>
11 #include <cstring>
12 #include <algorithm>
13
14 using namespace std;
15 const int n = 6;
16 const int M = 1005;
17 const int val[n] = {1, 5, 10, 20, 50, 100};
18 int x1, x2, x3;
19 int now, tot;
20 int i, j, k, a, b, suma, sumb, dis;
21 int sum[3], Cnt[n];
22 int d[2][M][M], cnt[3][n];
23
24 inline void update(int &x, int y){
25     if (x == -1) x = y;
26     else x = min(x, y);
27 }
28
29 inline int calc(){
30     return (abs(a - cnt[0][i]) + abs(b - cnt[1][i]) + abs(Cnt[i] - a - b - cnt[2][i])) / 2;
31 }
32
33 int main(){
34     scanf("%d%d%d", &x1, &x2, &x3);
35     for (i = 0; i < 3; ++i){
36         sum[i] = 0;
37         for (j = n - 1; j >= 0; --j){
38             scanf("%d", cnt[i] + j);
39             Cnt[j] += cnt[i][j];
40             sum[i] += cnt[i][j] * val[j];
41         }
42         tot += sum[i];
43     }
44
45     memset(d[0], -1, sizeof(d[0]));
46     d[0][sum[0]][sum[1]] = 0;
47     for (i = 0; i < n; ++i){
48         now = i & 1;
49         memset(d[now ^ 1], -1, sizeof(d[now ^ 1]));
50         for (j = 0; j <= tot; ++j){
51             for (k = 0; j + k <= tot; ++k){
52                 if (d[now][j][k] >= 0){
53                     update(d[now ^ 1][j][k], d[now][j][k]);
54                     for (a = 0; a <= Cnt[i]; ++a){
55                         for (b = 0; a + b <= Cnt[i]; ++b){
56                             suma = j + val[i] * (a - cnt[0][i]);
57                             sumb = k + val[i] * (b - cnt[1][i]);
58                             if (suma >= 0 && sumb >= 0 && suma + sumb <= tot){
59                                 dis = calc();
60                                 update(d[now ^ 1][suma][sumb], d[now][j][k] + dis);
61                             }
62                         }
63                     }
64                 }
65             }
66         }
67     }
68     int ea = sum[0], eb = sum[1], ec = sum[2];
69     ea += x3 - x1, eb += x1 - x2, ec += x2 - x3;
70     if (ea < 0 || eb < 0 || ec < 0 || ea + eb + ec != tot || d[n & 1][ea][eb] < 0)
71         printf("impossible\n");
72     else printf("%d\n", d[n & 1][ea][eb]);
73     return 0;
74 }