LA 7512 November 11th （思维漏洞）

题意：给定n*m个座椅，然后有b个是坏的，要做人，并且两个人不能相邻，问你最多坐多少人，最少坐多少人。

析：这个题其实并不难，只要当时一时没想清楚，结果就一直WA，就是最少的情况时，其实一个人可以占三个座位，而不是两个，就是这一点没想清楚，其他的就简单了。

代码如下：

#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f3f;
const double eps = 1e-8;
const int maxn = 1e6 + 5;
const int dr[] = {0, 0, -1, 1};
const int dc[] = {-1, 1, 0, 0};

int n, m, q;
vector<int> a[1010];

int main(){
    int T, ca = 0;
    scanf("%d", &T);
    while(T--){
        scanf("%d %d %d", &n, &m, &q);
        for(int i = 0; i <= n; ++i)  a[i].clear();
        int tmp1, tmp2;
        while(q--){
            scanf("%d %d", &tmp1, &tmp2);
            a[tmp1 + 1].push_back(tmp2 + 1);
        }
        for(int i = 1; i <= n; ++i)  sort(a[i].begin(), a[i].end());
        int ans1 = 0, ans2 = 0;
        for(int line = 1; line <= n; ++line){
            if(a[line].empty()){
                ans1 += (m+1) / 2;
                ans2 += (m+2) / 3;
                continue;
            }
            int head = 0, len = a[line].size();
            for(int i = 0; i < len; ++i){
                int lur = a[line][i];
                int cha = lur - head - 1;

                ans1 += (cha+1) / 2;
                ans2 += (cha+2) / 3;
                head = lur;
            }

            int lur = m + 1;
            int cha = lur - head - 1;
            ans1 += (cha+1) / 2;
            ans2 += (cha+2) / 3;
        }
        printf("Case #%d: %d %d\n", ++ca, ans1, ans2);
    }
    return 0;
}