P3842 [TJOI2007]线段
考察:线性dp
思路:
这道题就和POJ 1661差不多.f[i][0/1]表示到达第i条线段的0(左),1(右)的最小距离.
f[i][0] = min(f[i-1][0]+i-1条线左端点到i条线左端点的最短距离,f[i-1][1]+i-1条线右端点到第i条线左端点的最小值)
同理右端点.
1 #include <iostream> 2 #include <cstring> 3 using namespace std; 4 typedef pair<int,int> PII; 5 const int N = 20010; 6 int n,f[N][2]; 7 PII p[N]; 8 int getl(int x,int idx)//x到左点的最小值 9 { 10 int l = p[idx].first,r = p[idx].second,res = 0; 11 if(x<=l) res = 2*r-l-x; 12 else if(x>l&&x<r) res = min(x-l+(r-l)*2,r-x+r-l); 13 else res = x-l; 14 return res; 15 } 16 int getr(int x,int idx)//x到右端点的最小值. 17 { 18 int l = p[idx].first,r = p[idx].second,res = 0; 19 if(x<=l) res = r-x; 20 else if(x>l&&x<r) res = min(x-l+r-l,r-x+(r-l)*2); 21 else res = x+r-2*l; 22 } 23 int solve() 24 { 25 p[0].first = 1,p[0].second = 1; 26 for(int i=1;i<=n;i++) 27 { 28 int l = p[i-1].first,r = p[i-1].second; 29 f[i][0] = min(f[i-1][0]+getl(l,i),f[i-1][1]+getl(r,i)); 30 f[i][1] = min(f[i-1][0]+getr(l,i),f[i-1][1]+getr(r,i)); 31 if(i!=1) f[i][0]++,f[i][1]++; 32 } 33 int l = p[n].first,r = p[n].second; 34 int ans = min(f[n][0]+abs(n-l),f[n][1]+abs(n-r)); 35 return ans; 36 } 37 int main() 38 { 39 scanf("%d",&n); 40 for(int i=1;i<=n;i++) scanf("%d%d",&p[i].first,&p[i].second); 41 int res = solve(); 42 printf("%d\n",res); 43 return 0; 44 }
