USACO: Barn Repair

来源：互联网发布：软件项目经理的职责编辑：程序博客网时间：2024/06/03 20:39

/*PROG: barn1LANG: C++11*/#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <cstdlib>#include <climits>#include <ctype.h>#include <queue>#include <stack>#include <vector>#include <utility>#include <deque>#include <set>#include <map>#include <iostream>#include <fstream>#include <algorithm>using namespace std;#define mst(a,b) memset(a,b,sizeof(a))#define eps  10e-8typedef long long ll;const int N = 100010;const ll MOD = 1000000007;const int INF = 0x7fffffff;int m ,s ,c, k, i;bool stalls[500];int main(){    ifstream fin ("barn1.in");    ofstream fout ("barn1.out");        fin >> m >> s >> c;    for(i = 0; i < c; i++){        fin >> k;        stalls[k] = 1;    }    // get rid of both end    i = 1;    while(i <= s && !stalls[i] ){        i++;    }    vector<int> vacancy;    int empty = 0;// only when empty slots is between occupied stalls    int board =  0;    while(1){        empty = 0;        while(i <= s && !stalls[i]){            i++;            empty++;        }        if( i <= s){            board++;            if(empty)                vacancy.push_back(empty);        }        else {            break;        }        while( i <= s && stalls[i]){            i++;        }        if(i == s+1){            break;        }    }    sort(vacancy.begin(), vacancy.end());        for(int i = 0; i < board -m; i++){        c += vacancy[i];    }    fout << c << endl;        fin.close();    fout.close();    return 0;}

贪心：首先计算没有木板限制时为达到总length最短所需要的木板。用合并的方式来达到减少木板的目的。每一次合并浪费掉的木板最短，结果就是最优解。

具体的一次扫描，计算初始木板个数的同时统计连续无奶牛的stall的长度，最后排序，从短的开始合并就可以了。

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