老BOJ 03 Guess

Guess

Accept:357    Submit:2244Time Limit:3000MS    Memory Limit:65536KBDescription
A competition was just over. It had 3 problems and n players. Each player had an ID number from 1 to n. The final rank was decided by the total score of the 3 problems. The higher the total score was, the higher a player ranked (the smaller the rank number). If two players got the same total score, the one with the smaller ID number got a higher rank. We’ve known for each problem, how much score each player might get if he din’t solve totally wrong (if solved totally wrong, the player got zero in the problem). However, we don’t know whether a player did get score in a problem. For a predicted final rank, you need to judge if the rank is possible.

Input
Input contains several cases. For each case, the first line is an integer n, (n <= 16384) to indicate the number of players, followed by n lines, the ith of which contains three real numbers a, b, c (0<=a, b, c < 1000. a, b and c have 2 decimal places at most.) to respectively indicate the score of each problem Player i might get if he didn’t solve totally wrong. Another line containing n integers follows to indicate the player ID number in the order from rank 1st to rank nth .
The last case is followed by a line containing only a zero.

Output
For each case, if the rank is possible, output the highest possible total score for the player with the lowest rank (calculate to 2 decimal places), otherwise output “No solution" (quotes for clarity).

Sample Input

3
100 200 300
100 200 300
100 200 300
1 2 3
3
100 200 300
100 200 300
100 200 300
3 2 1
0


Sample Output

Case 1: 600.00
Case 2: 400.00


Source

ACM Beijing 2006-E

贪心+枚举、只需要求出所有的得分可能，然后由高到低进行找出即可

INX_MAX存在于<climits>

#include<cstdlib>#include<algorithm>#include<cstdio>#include<iostream>#include<cmath>#include<climits>using namespace std;int main(){    int n,count(0);    while(scanf("%d",&n)&& n)    {        ++count;        int* rank =new int[n+5];//rank数组：rank[i]表示第i+1位置的人的序号        int** score =new int*[n];        bool fail =false;        int lastscore(INT_MAX),lastone(INT_MAX),temp;//上一位选手的名次和序号，取一个较大的值用来覆盖        float f1,f2,f3;        for(int i(0); i != n; ++i)        {            score[i]=new int[8];            scanf("%f%f%f",&f1,&f2,&f3);            score[i][0]=(int)((f1+0.001)*100);            score[i][1]=(int)((f2+0.001)*100);            score[i][2]=(int)((f3+0.001)*100);            score[i][3]= score[i][0]+score[i][1];            score[i][4]= score[i][1]+score[i][2];            score[i][5]= score[i][0]+score[i][2];            score[i][6]= score[i][0]+score[i][1]+score[i][2];            score[i][7]=0;            sort(score[i],score[i]+8);//这位选手可能得到的分数，共8种        }        for(int i(0); i != n; ++i)            scanf("%d",&rank[i]);        for(int i(0); i != n; ++i)        {            int j = rank[i]-1;            int begin =7;            while(begin >=0&&(score[j][begin]> lastscore ||(score[j][begin]== lastscore  && j < lastone)))//判断过程，从最大的分数开始：如果不符合要求，则替换为下一个分数，直到begin<0                begin--;            if(begin <0)//说明不存在满足条件的分数            {                fail =true;                break;            }            else//否则继续测试下一位选手            {                lastone = j;                lastscore = score[j][begin];            }        }        if(!fail)            printf("Case %d: %.2lf\n",count,(double)lastscore/100);        else printf("Case %d: No solution\n",count);        delete[] rank;        for(int i(0); i != n; ++i)            delete[] score[i];        delete[] score;    }    return 0;}