HDU 4004

来源：互联网发布：库里职业生涯数据统计编辑：程序博客网时间：2024/06/09 21:24

二分可做。

只要是能确定答案域的，且可通过验证函数缩小答案域的，基本上都可能可以用二分做。

#include<cstdio>#include<cstdlib>int cmp(const void* a,const void* b){    return *(int*)a-*(int*)b;}bool check(int len,int n,int m,int* stone,int temp){    int go=0,k=0,times=0;    while(go<len)    {        //printf("go=%d len=%d stone[%d]=%d times=%d\n",go,len,k,stone[k],times);        while(go+temp>=stone[k]&&k<n+1) k++;        if(go+temp>=len&times<m) return true;        else if(go+temp>=stone[k-1]&&go<stone[k-1]&times<m) {go=stone[k-1]; times++;}        else return false;    }}int slove(int len,int n,int m,int* stone){    int left=0,right=len;    while(left<right)    {        int mid=(left+right)/2;        //printf("mid=%d\n",mid);        if(check(len,n,m,stone,mid)) right=mid;        else left=mid+1;        //printf("left=%d right=%d\n",left,right);    }    return right;}int main(){    int l,n,m;    while(scanf("%d%d%d",&l,&n,&m)!=EOF)    {        int stone[n+1];        for(int i=0;i<n;i++)            scanf("%d",&stone[i]);        stone[n]=l;        qsort(stone,n,sizeof(stone[0]),cmp);        printf("%d\n",slove(l,n,m,stone));    }    return 0;}

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