POJ 1015 Jury Compromise
传送门：http://vjudge.net/problem/POJ-1015
题目翻译
N个人，有P参数和D参数。从中选m个人，使得|∑P – ∑D|最小（一样则优先选取∑P + ∑D大的）
代码
F[i][K]表示已经选了i个人，∑P – ∑D = K 的最大 ∑P + ∑D
然后记录一下方案。
题解
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#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int Maxn = 200; const int Maxm = 20; int n, m; struct JuryNode { int def,pro; }; JuryNode jury[Maxn + 5]; int DP[Maxm + 5][Maxn * Maxm * 2 + 5]; int pre[Maxm + 5][Maxn * Maxm * 2 + 5]; inline int absX(int x) { if (x < 0) return -x; return x; } struct Node { int v; int prev; Node(){} Node(int v0, int p0) { v = v0; prev = p0; } }; Node nn[Maxn * Maxm * Maxn * 10 + 5]; int tot = 0; int pList[Maxm + 5]; void print(int x) { if (x==0) { printf("\n"); return ; } printf("%d ",nn[x].v); print(nn[x].prev); } inline bool cmp(int a, int b) { return a < b; } inline void solve(int T) { for (int i = 1; i <= n; i++) scanf("%d%d", &jury[i].pro, &jury[i].def); tot = 0; memset(DP, -1, sizeof(DP)); memset(pre, 0 ,sizeof(pre)); int maxDelta = n * Maxm; for (int i = 1; i <= n; i++){ for (int j = min(i, m) - 1; j >= 0; j--){ if (j == 0) { if (DP[j + 1][jury[i].pro - jury[i].def + maxDelta] == -1 || DP[j + 1][jury[i].pro - jury[i].def + maxDelta] < jury[i].def + jury[i].pro) { DP[j + 1][jury[i].pro - jury[i].def + maxDelta] = jury[i].def + jury[i].pro; nn[++tot] = Node(i, 0); pre[j + 1][jury[i].pro - jury[i].def + maxDelta] = tot; /*printf("Juries : "); print(tot); int sum = DP[j + 1][jury[i].pro - jury[i].def + maxDelta], minu = jury[i].pro - jury[i].def ; printf("--Pro : %d Def : %d\n", (sum + minu) / 2, (sum - minu) / 2);*/ } } else { for (int delta = 0; delta <= maxDelta * 2; delta++) { if (DP[j][delta] != -1) { if (DP[j + 1][delta + jury[i].pro - jury[i].def] == -1 || DP[j + 1][delta + jury[i].pro - jury[i].def] < DP[j][delta] + jury[i].pro + jury[i].def){ DP[j + 1][delta + jury[i].pro - jury[i].def] = DP[j][delta] + jury[i].pro + jury[i].def; nn[++tot] = Node(i, pre[j][delta]); pre[j + 1][delta + jury[i].pro - jury[i].def] = tot; /*printf("Juries : "); print(tot); int sum = DP[j + 1][delta + jury[i].pro - jury[i].def], minu = delta + jury[i].pro - jury[i].def - maxDelta; printf("--Update : [%d][%d] Pro : %d Def : %d\n", j + 1, delta + jury[i].pro - jury[i].def - maxDelta,(sum + minu) / 2, (sum - minu) / 2);*/ } } } } } } int Ans = maxDelta,ansSum = 0,ansMinus = 0,preHead = 0; for (int delta = 0; delta <= maxDelta * 2; delta++) { if (DP[m][delta] != -1) { //printf("Delta Find %d!\n",delta); if (Ans > absX(delta - maxDelta) || (Ans == absX(delta - maxDelta) && DP[m][delta] > ansSum)) { Ans = absX(delta - maxDelta); ansSum = DP[m][delta]; ansMinus = delta - maxDelta; preHead = pre[m][delta]; } } } for (int i = 1; i <= m; i++) { pList[i] = nn[preHead].v; preHead = nn[preHead].prev; } sort(pList + 1, pList + m + 1, cmp); printf("Jury #%d\nBest jury has value %d for prosecution and value %d for defence:\n", T, (ansSum + ansMinus) / 2, (ansSum - ansMinus) / 2); for (int i = 1; i <= m; i++) printf(" %d", pList[i]); printf("\n"); } int main() { int T = 0; while(scanf("%d%d", &n, &m) != EOF){ if (n == 0 && m == 0) return 0; solve(++T); } return 0; } |
原文链接：POJ 1015 Jury Compromise
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