poj 1276 Cash Machine

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题目大意：给定一个数m,还有一组数，分别表示这个元素的个数num[i], 单个价值为value[i]。要你求最这组元素价值和最接近m的最大值

解题思路：多重背包问题，拆分2进制或单调队列

同poj 1014一样

http://blog.csdn.net/xiaoxiaoluo/article/details/7816378

2进制拆分方法：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 100010;const int maxm = 10010;int dp[maxn];int cash, n, index, nk[11], dk[11], value[maxm];void spilt(int n, int v);int main(){while(scanf("%d %d", &cash, &n) != EOF){memset(dp, 0, sizeof(dp));index = 0;for(int i = 0; i < n; i++){scanf("%d %d", &nk[i], &dk[i]);spilt(nk[i], dk[i]);}if(cash == 0){printf("0\n");continue;}for(int i = 0; i < index; i++){for(int j = cash; j >= value[i]; j--)dp[j] = max(dp[j], dp[j - value[i]] + value[i]);}printf("%d\n", dp[cash]);}return 0;}void spilt(int n, int v){int x, i = 0, tmp = 0;while(true){x = 1 << i;if(tmp + x > n)break;tmp += x;value[index++] = x * v;i++; }x = n - tmp;if(x > 0)value[index++] = x * v;}

单调队列的方法：

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;struct node{int index, value;};const int maxn = 100010;const int maxm = 10010;int dp[maxn];int cash, n, nk[11], dk[11];node que[maxn];int main(){while(scanf("%d %d", &cash, &n) != EOF){memset(dp, 0, sizeof(dp));for(int i = 0; i < n; i++)scanf("%d %d", &nk[i], &dk[i]);if(cash == 0){printf("0\n");continue;}for(int i = 0; i < n; i++){int t = nk[i] * dk[i];if(nk[i] == 1){for(int j = cash; j >= dk[i]; j--)dp[j] = max(dp[j], dp[j - dk[i]] + dk[i]);}else if(t >= cash){for(int j = dk[i]; j <= cash; j++)dp[j] = max(dp[j], dp[j - dk[i]] + dk[i]);}else{for(int j = 0; j < dk[i]; j++){int head = 0, tail = -1;for(int k = j, nu = 0; k <= cash; k += dk[i], nu++){if(head <= tail && k - que[head].index > t)head++;int tt = dp[k] - nu * dk[i];while(head <= tail && que[tail].value < tt)tail--;tail++;que[tail].index = k;que[tail].value = tt;dp[k] = que[head].value + nu * dk[i];}}}}printf("%d\n", dp[cash]);}return 0;}