BZOJ 3892 [Usaco2014 Dec]Marathon 动态规划
2015-02-27 20:00
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题目大意:给定n个点,定义从一个点到另一个点的距离为曼哈顿距离,要求从点1依次走到点n,中途可以跳过k个点不走,求最小距离和
令f[i][j]表示从第一个点走到第i个点中途跳过j次的最小距离和
则有f[i][j]=min{f[i-k-1][j-k]+dis[i-k-1][i]}
时间复杂度O(n^3)
令f[i][j]表示从第一个点走到第i个点中途跳过j次的最小距离和
则有f[i][j]=min{f[i-k-1][j-k]+dis[i-k-1][i]}
时间复杂度O(n^3)
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define M 510 using namespace std; struct Point{ int x,y; friend istream& operator >> (istream &_,Point &p) { scanf("%d%d",&p.x,&p.y); return _; } friend int Distance(const Point &p1,const Point &p2) { return abs(p1.x-p2.x)+abs(p1.y-p2.y); } }points[M]; int n,k,f[M][M]; int main() { int i,j,k; cin>>n>>::k; for(i=1;i<=n;i++) cin>>points[i]; memset(f,0x3f,sizeof f); f[1][0]=0; for(i=2;i<=n;i++) for(j=0;j<=i-2&&j<=::k;j++) for(k=0;k<=j;k++) f[i][j]=min(f[i][j],f[i-k-1][j-k]+Distance(points[i-k-1],points[i])); cout<<f [min(n-2,::k)]<<endl; return 0; }
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