[Usaco 2011 Dec]Umbrellas for Cows
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题目大意:
FJ的N ( 1 < = N < = 5,000)头奶牛不喜欢被淋湿。它们站在数轴上(1到M ( 1 < = M < = 100,000 )),现在FJ要买伞,遮住i个点的伞要Ci元,问给所有的牛遮雨要花多少钱?
输入数据第1行:两个用空格隔开的整数: N和M
这题就是一道裸dp,先按坐标排序 然后记f[i]表示覆盖前i个牛需要多少钱 f[i]=min{f[j-1]+cost[a[i]-a[j]+1]) 1<=j<=i
附代码
#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>const int maxn=5010,maxm=100010;using namespace std;#ifdef judgeonlineconst int maxchar=20000000;char buf[maxchar],*wbuf=buf;inline int getint(){int tmp=0;while (*wbuf<'0' || *wbuf>'9') ++wbuf;for (;*wbuf>='0' && *wbuf<='9';++wbuf) tmp=tmp*10+*wbuf-'0';return tmp;}#elseinline int getint(){ char ch=getchar(); int tmp=0; while (ch<'0' || ch>'9') ch=getchar(); for (;ch<='9' && ch>='0';ch=getchar()) tmp=tmp*10+ch-'0'; return tmp;}#endifint n;int wh[maxn];int price[maxm];void init(){n=getint();int m=getint();for (int i=1;i<=n;++i)wh[i]=getint();for (int i=1;i<=m;++i)price[i]=getint();for (int i=m-1;i>=1;--i)price[i]=min(price[i],price[i+1]);}int f[maxn];void dp(){sort(wh+1,wh+n+1);memset(f,63,sizeof(f));f[0]=0;for (int i=1;i<=n;++i)for (int j=1;j<=i;++j)f[i]=min(f[i],f[j-1]+price[wh[i]-wh[j]+1]);printf("%d\n",f[n]);}int main(){init();dp();return 0;}
Ps:这题有个坑爹的地方,就是有可能越长越便宜,然后就Wa了2次。。。
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