Although only a fraction of this topic of 1500, but I still feel very thought, I did not write it today, then I looked at the problem solution
Rarely do this interval coverage of the subject, is not very good, then talk about the idea after I read the explanations.
Title effect that: give you a number of columns, the number of columns 0 means the place no heaters, 1 representatives, each heater is a range [i-r + 1, i + r-1]
Ask you to use the least number of heaters for the whole series are heated.
We can have a heater into a queue, and with a number represented haves [. 1, haves] range has been covered, then use a few now be expressed according to a point
This is a position the farthest point you can choose, if can not find the farthest point of it means not meet the requirements of coverage.
#include <cstring> #include <queue> #include <cstdlib> #include <cstdio> #include <iostream> #include <string> #include <bitset> #include <algorithm> #include <map> #include <vector> #define inf 0x3f3f3f3f #define inf64 0x3f3f3f3f3f3f3f3f using namespace std; typedef long long ll; const int maxn = 1e5 + 10; int main() { int n, r; queue<int>que; scanf("%d%d", &n, &r); for(int i=1;i<=n;i++) { int x; scanf("%d", &x); if (x) que.push(i); } int haves = 0, now = r, ans = 0; while(!que.empty()) { int u = que.front(); que.pop(); if (haves >= n) break; if (u > now) { printf("-1\n"); return 0; } if(!que.empty()) { int v = que.front(); if (v <= now) continue; ans++; haves = u + r - 1; now = u + 2 * r - 1; } else { ans++; haves = u + r - 1; now = u + 2 * r - 1; } } if (haves >= n) printf("%d\n", ans); else printf("-1\n"); return 0; }