topcoder SRM 624 DIV2 BuildingHeightsEasy

OpenSoucre發表於2014-06-13

從大到小遍歷一遍,每次取M個元素,然後求得最小的floor即可

    int minimum(int M, vector <int> heights)
    {
        sort(heights.begin(),heights.end());
        int minFloor = 10000;
        for(int i = heights.size()-1; i >=M-1; -- i){
            int floor = 0;
            for(int j = 0; j < M; ++j) floor +=heights[i]-heights[i-j];
            minFloor =min(minFloor,floor);
        }
        return minFloor;
    }

  

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