POJ 1015 - Jury Compromise

题目的意思是，从n个数对<Pi, Di>中选出m个，要求：

1. |Sum{Pi} - Sum{Di}|最小

2. 满足1的情况下，Sum{Pi + Di}最大

并且输出这m个的编号。

这里n<=200, m<=20, 0<=Pi,Di<=20

解法：三维动态规划。设s[i] = Pi - Di, t[i] = Pi + Di,

f[i][j][k]表示从前i个人中选出j个人，满足它们的s值之和等于k，最大的t值之和，如果无解则等于-1。问题的解就是满足f[n][m][k]有解的绝对值最小的k。

动态转移方程：

对应第i个人选和不选两种情况，f[i][j][k] = max { f[i-1][j][k], f[i-1][j-1][k-s[i]]+t[i] }

其中1<=i<=n, 1<=j<=min(i,20), -20*j<=k<=20*j

初始条件：f[i][0][0]=0, 其他都为-1.

注意考虑f无解的情况，转移时需要做一些条件判断。

由于题目还要求记录下m个人具体的编号，所以还需要一个三维数组path[i][j][k]记录对应f[i][j][k]的最后一个人编号，最后沿着path回退即可构造答案序列。

需要注意的是k有可能是负值，所以实现时要将f的第三维度都加上400，保证下标是正数。

附上AC代码：

#include <iostream>#define MIN(x, y) ((x) < (y) ? (x) : (y))using namespace std;int s[201], t[201], f[201][21][801], path[201][21][801], prosecution[201], defense[201];int main(){int n, m, nCase;for (nCase = 1;; nCase++){cin >> n >> m;if (n == 0 && m == 0)break;for (int i = 1; i <= n; i++){cin >> prosecution[i] >> defense[i];s[i] = prosecution[i] - defense[i];t[i] = prosecution[i] + defense[i];}/* f[i][0][0] = 0, all the others = -1 */for (int i = 0; i <= 200; i++){for (int j = 0; j <= 20; j++){for (int k = 0; k <= 800; k++){if (j == 0 && k == 400)f[i][j][k] = 0;else{f[i][j][k] = -1; /* no solution */}}}}for (int i = 1; i <= n; i++){for (int j = 1; j <= MIN(i, 20); j++){for (int k = -20 * j; k <= 20 * j; k++){k += 400;/* not select */if (f[i - 1][j][k] != -1){f[i][j][k] = f[i - 1][j][k];path[i][j][k] = path[i - 1][j][k];}/* select */if (f[i - 1][j - 1][k - s[i]] != -1 && f[i - 1][j - 1][k - s[i]] + t[i] > f[i][j][k]){f[i][j][k] = f[i - 1][j - 1][k - s[i]] + t[i];path[i][j][k] = i;}k -= 400;}}}for (int i = 0; i <= 400; i++){if (f[n][m][400 - i] != -1 || f[n][m][400 + i] != -1){int optK = (f[n][m][400 - i] > f[n][m][400 + i] ? 400 - i : 400 + i);int ans[21], sum_prosecution = 0, sum_defense = 0;/* rebuild the path */int ii = n, jj = m, kk = optK;for (int j = 1; j <= m; j++){ans[j] = path[ii][jj][kk];ii = ans[j] - 1;jj--;kk -= s[ans[j]];sum_prosecution += prosecution[ans[j]];sum_defense += defense[ans[j]];}/* write result */cout << "Jury #" << nCase << endl;cout << "Best jury has value " << sum_prosecution << " for prosecution and value " << sum_defense << " for defence:" << endl;/* write path */for (int j = m; j >= 1; j--)cout << ' ' << ans[j];cout << endl << endl;break;}}}return 0;}