Codeforces 567D One-Dimensional Battle Ships

One-Dimensional Battle Ships

题目链接：

http://codeforces.com/problemset/problem/567/D

解题思路：

First, we should understand when the game ends. It will happen when on the n-sized board it will be impossible to place k ships of size a. For segment with length len we could count the maximum number of ships with size a that could be placed on it. Each ship occupies a + 1cells, except the last ship. Thus, for segment with length len the formula will look like  (we add "fictive" cell to len cells to consider the last ship cell). This way, for [l..r] segment the formula should be .

For solving the problem we should store all the [l..r] segments which has no ''free'' cells (none of them was shooted). One could use (std: : set) for that purpose. This way, before the shooting, there will be only one segment [1..n]. Also we will store current maximum number of ships we could place on a board. Before the shooting it is equal to .

With every shoot in cell x we should find the segment containing shooted cell (let it be [l..r]), we should update segment set. First, we should delete [l..r] segment. It means we should decrease current maximum number of ships by  and delete it from the set. Next, we need to add segments [l..x - 1] and [x + 1..r] to the set (they may not be correct, so you may need to add only one segments or do not add segments at all) and update the maximum number of ships properly. We should process shoots one by one, and when the maximum number of ships will become lesser than k, we must output the answer. If that never happen, output  - 1.

AC代码：

#include <iostream>
#include <cstdio>
#include <set>
using namespace std;

set<int> s;

int main(){
    int n,k,a,m;
    while(~scanf("%d%d%d",&n,&k,&a)){
        scanf("%d",&m);
        s.clear();
        s.insert(0);
        s.insert(n+1);
        int ans = (n+1)/(a+1);
        int ans_id = -1;
        for(int i = 1; i <= m; i++){
            int x;
            scanf("%d",&x);
            if(ans < k)
                continue;
            s.insert(x);
            set<int>::iterator it = s.find(x);
            int l = *(--it);
            int r = *(++(++it));
            ans -= (r-l)/(a+1);
            ans += (x-l)/(a+1) + (r-x)/(a+1);
            if(ans < k)
                ans_id = i;
        }
        printf("%d\n",ans_id);
    }
    return 0;
}