poj2392（多重背包dp，可行性类型O（nm）复杂度的解法。）

/*translation:有k种石块，每种石块有ci个，每种石块所在的高度不能超过ai，求用这k种石头所能达到的最高高度solution:多重背包dp可行性解法，贪心先排序，然后按照多重背包dp可行性解法即可acnote:此题用常规的多重背包wa了，不知道为啥？date:2016.9.8*/#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 400 +5;struct Stock {int h, a, c;//石块的高度，高度上限，该类石块的数量bool operator < (const Stock& rhs) const {if(a == rhs.a)return h > rhs.h;elsereturn a < rhs.a;}} stocks[60000];int k;int dp[40000 + 5];int main(){//freopen("in.txt", "r", stdin);    while(~scanf("%d" ,&k)) {int max_h = -1;for(int i = 0; i < k; i++) {scanf("%d%d%d", &stocks[i].h, &stocks[i].a, &stocks[i].c);max_h = max(max_h, stocks[i].a);}sort(stocks, stocks + k);memset(dp, -1, sizeof(dp));dp[0] = 0;for(int i = 0; i < k; i++) {for(int j = 0; j <= stocks[i].a; j++) {if(dp[j] >= 0) {dp[j] = stocks[i].c;} else if(j < stocks[i].h || dp[j - stocks[i].h] <= 0) {dp[j] = -1;} else {dp[j] = dp[j - stocks[i].h] - 1;}}}int ans;for(int i = 0; i <= stocks[k - 1].a; i++)if(dp[i] >= 0)ans = i;printf("%d\n", ans);    }    return 0;}