1079. Total Sales of Supply Chain

Tips:

• 真是no zuo no die.说好的尽量用C来写,脑子一抽,又用Java来写了,最直接的方法,结果来了个超时.好吧,以为是算法问题,继续改,各种思路,仍得不到解决.换C++吧,还是各种思路,没能AC.最后想试试看用C++来写最开始的最直接的方法,呵呵,AC了…再说一遍,没什么事别用Java!!!
• 这题没什么技巧,也没什么陷阱,无非就先把输入的数据用二维数组保存起来,然后用到了队列.其它没什么好说的了.虽然是个25分的题,但我认为难度不大.

#include "iostream"using namespace std;int main(int argc, const char * argv[]) {    int n,i,j,m,len,index;    double p,r,amount;    int **chain,*tree;    double *price;    while (scanf("%d %lf %lf",&n,&p,&r) != EOF) {        chain = new int*[n];        tree = new int[n];        price = new double[n];        price[0] = p;        amount = 0.0;        for(i = 0;i < n;i++){            scanf("%d",&m);            if(m == 0)                len = 2;            else                len = m + 1;            chain[i] = new int[len];            chain[i][0] = m;            for(j = 1;j < len;j++)                scanf("%d",&chain[i][j]);        }        tree[0] = 0;        for(i = 0,index = 1;i < n;i++){            if(chain[tree[i]][0] == 0)                amount += price[tree[i]] * chain[tree[i]][1];            else{                for(j = 1;j <= chain[tree[i]][0];j++){                    tree[index++] = chain[tree[i]][j];                    price[chain[tree[i]][j]] = price[tree[i]] * (1 + r / 100);                }            }        }        printf("%.1lf\n",amount);    }    return 0;}