Codeforces Round #600 (Div. 2) E. Antenna Coverage(dp)
题意:
m只蚂蚁,每只蚂蚁的位置是\(x_i\),分数是\(s_i\),覆盖范围是\([x_i - s_i; x_i + s_i]\),每个硬币可以使一直蚂蚁的\(s_i\)+1,求覆盖整个\([1;m]\)的最少硬币
思路:
\(f[pos][0]\)表示\([1,pos]\)没有被覆盖还要花费的最少硬币,\(f[pos][1]\)表示\([1,pos]\)被覆盖的最小花费硬币,对于每个\(pos\)枚举蚂蚁,有以下这几种情况
- \(pos\)在\(x_i\)的右边,且蚂蚁\(i\)扩展到\(pos\)时的左边已经有蚂蚁覆盖,则蚂蚁\(i\)扩展到\(pos\)时\(f[pos][1] = min(f[pos][1],f[max(2*x[j]-pos-1,0)][1]+max(pos-x[j]-s[j],0));\)
- \(pos\)在\(x_i\)的右边,且蚂蚁\(i\)扩展到\(pos\)时的左边无蚂蚁覆盖,则蚂蚁\(i\)扩展到\(pos\)时\(f[pos][1] = min(f[pos][1],f[max(2*x[j]-pos-1,0)][0]+1+max(pos-x[j]-s[j],0));\)
-
\(pos\)在\(x_i\)的左边,则只需扩展\(x_i\)的左侧\(f[pos][1] = min(f[pos][1],min(f[max(x[j]-s[j]-1,0)][1],f[max(x[j]-s[j],0)][0]));\)
\(f[i][1]\)和\(f[i][0]\)的区别是若左边已经有蚂蚁覆盖了,就可以少去覆盖\([i-1,i]\)的花费
代码:
#include <bits/stdc++.h>
using namespace std;
const int N = 200100;
int f[N][2],s[N],x[N];
int main()
{
int n,m;
cin >> n >> m;
for (int i = 1; i <= m; i++) f[i][0]=i-1,f[i][1]=1<<30;
f[0][1] = 0;
for (int i = 1; i <= n; i++)
{
cin >> x[i] >> s[i];
}
for (int i = 1; i<= m; i++)
for (int j = 1; j <= n;j++)
if ( i >= x[j])
{
f[i][1] = min(f[i][1],f[max(2*x[j]-i-1,0)][1]+max(i-x[j]-s[j],0));
f[i][1] = min(f[i][1],f[max(2*x[j]-i-1,0)][0]+1+max(i-x[j]-s[j],0));
}
else
{
f[i][1] = min(f[i][1],f[max(x[j]-s[j]-1,0)][1]);
f[i][1] = min(f[i][1],f[max(x[j]-s[j],0)][0]);
}
cout << f[m][1] << endl;
}