https://vjudge.net/problem/UVA-1632
题意:
直线上有n个点,其中第i个点的坐标是xi,且它会在di秒之后消失。Alibaba可以从任意位置出发,求访问完所有点的最短时间。
思路:
区间DP。
d[i][j][0]用来表示访问完区间 i ~ j 之间所有点的最短时间,并且此时处于 i 点,相反的,d[i][j][1]表示的是处于 j 点。
1 #include<iostream> 2 #include<string> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 7 const int maxn = 10000 + 5; 8 const int INF = 10000000; 9 10 int a[maxn], b[maxn]; 11 int n; 12 int d[maxn][maxn][2]; 13 14 int main() 15 {17 while (cin >> n) 18 { 19 memset(d, 0, sizeof(d)); 20 for (int i = 1; i <= n; i++) 21 cin >> a[i] >> b[i]; 22 for (int i = n; i >=1; i--) 23 { 24 for (int j = i + 1; j <= n; j++) 25 { 26 d[i][j][0] = min(d[i + 1][j][0] + a[i + 1] - a[i], d[i + 1][j][1] + a[j] - a[i]); 27 if (d[i][j][0] >= b[i]) d[i][j][0] = INF; 28 d[i][j][1] = min(d[i][j-1][1] + a[j] - a[j - 1], d[i][j - 1][0] + a[j] - a[i]); 29 if (d[i][j][1] >= b[j]) d[i][j][1] = INF; 30 } 31 } 32 int ans = min(d[1][n][0], d[1][n][1]); 33 if (ans == INF) puts("No solution"); 34 else printf("%d ", ans); 35 } 36 return 0; 37 }