Bzoj1021 [shoi2008] debt cyclic debt

It seems that I have heard of this question last summer... At that time, YY had a plane with three axes, and the angle between axes/3...

Correct solution... Of course DP.

F [I] [J] [k] indicates the I-th nominal value. The first person has J yuan, and the second person has a minimum exchange of K yuan.

So it's just a question about the backpack. Because the DP equation is too complicated, please refer to the program...

(There is also a very powerful pruning program I did not add, because it is too lazy ≥ v ≤ ...... Success! Don't hit me ~~~ 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 }

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Bzoj1021 [shoi2008] debt cyclic debt