标签：ber char less define name 传送门 style from pid
题意：it will be a strictly decreasing sequence from the 1st maximum , 2nd maximum .. to the K-th maximum. If the total number of different values is less than K,just ouput 0.
输入：T组数据 体积V 要求的K N , V, K(N <= 100 , V <= 1000 , K <= 30)
各项价值
各项体积
定义dp[i][j][k]为利用前i个和体积j能获得的不重复的第K大值。类似于01背包，可以用二维的dp[j][k]去存储信息。当取第i个物品时，枚举取与不取，得到体积j下的2k种情况，选择其中的k种更新答案即可。就像我有两个数列，要求第K大值，只要分别得知两个数列里的前K个值即可
复杂度是O(NVK)
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 #define INF 0x3f3f3f3f 6 #define MOD 1000000007 7 using namespace std; 8 typedef long long LL; 9 10 const int maxn = 1e2 + 10; 11 const int maxv = 1e3 + 10; 12 const int maxk = 35; 13 14 int dp[maxv][maxk]; 15 int slt[maxk], no_slt[maxk]; 16 int val[maxn], vol[maxn]; 17 18 int N, V, K; 19 20 int main(int argc, const char * argv[]) { 21 int T; scanf("%d", &T); 22 while (T--) { 23 memset(dp, 0, sizeof(dp)); 24 scanf("%d%d%d", &N, &V, &K); 25 for (int i = 0; i < N; i++) scanf("%d", &val[i]); 26 for (int i = 0; i < N; i++) scanf("%d", &vol[i]); 27 for (int i = 0; i < N; i++) { 28 for (int j = V; j >= vol[i]; j--) { 29 for (int k = 1; k <= K; k++) { 30 slt[k] = dp[j - vol[i]][k] + val[i]; 31 no_slt[k] = dp[j][k]; 32 } 33 slt[K + 1] = -1; no_slt[K + 1] = -1; 34 int p1 = 1, p2 = 1, p3 = 1; 35 while (p1 <= K && (slt[p2] != -1 || no_slt[p3] != -1)) { 36 if (slt[p2] > no_slt[p3]) { 37 dp[j][p1] = slt[p2++]; 38 } else { 39 dp[j][p1] = no_slt[p3++]; 40 } 41 if (dp[j][p1] != dp[j][p1 - 1]) 42 p1++; 43 } 44 45 } 46 } 47 printf("%d\n", dp[V][K]); 48 } 49 return 0; 50 }
标签：ber char less define name 传送门 style from pid
原文地址：http://www.cnblogs.com/xFANx/p/7414266.html
