NOIP真题题解（二）二分

1. NOIP2010提高组 关押罪犯

来源：NOIP2003提高组 https://ac.nowcoder.com/acm/contest/258/C

(二分，染色法判断二分图) O((N+M)logC)

• 将罪犯当做点，罪犯之间的仇恨关系当做点与点之间的无向边，边的权重是罪犯之间的仇恨值。
  那么原问题变成：将所有点分成两组，使得各组内边的权重的最大值尽可能小。
  我们在$[0,10^9]$ 之间枚举最大边权$limist$, 当$limist$固定之后，剩下的问题就是：
  • 判断能否将所有点分成两组，使得所有权值大于 limit 的边都在组间，而不在组内。也就是判断由所有点以及所有权值大于 limit 的边构成的新图是否是二分图。
• 判断二分图可以用染色法，时间复杂度是$O(N+M)$, 其中 N 是点数，M 是边数，可以参考AcWing 860. 染色法判定二分图。
  

#include 
#include 
#include 
#include 
using namespace std; const int N = 20010,
   M = 200010;

int n, m;
int h[N], e[M], w[M], ne[M], idx;
int color[N];

void add(int a, int b, int c)
{
   e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}

bool dfs(int u, int c, int limit)
{
   color[u] = c;
   for (int i = h[u]; ~i; i = ne[i])
   {
       if (w[i] <= limit) continue;
       int j = e[i];
       if (color[j])
       {
           if (color[j] == c) return false;
       }
       else if (!dfs(j, 3 - c, limit)) return false;
   }

   return true;
}

bool check(int limit)
{
   memset(color, 0, sizeof color);

   for (int i = 1; i <= n; i++)
       if (color[i] == 0)
           if (!dfs(i, 1, limit))
               return false;
   return true;
}

int main()
{
   scanf("%d%d", &n, &m);

   memset(h, -1, sizeof h);
   while (m--)
   {
       int a, b, c;
       scanf("%d%d%d", &a, &b, &c);
       add(a, b, c);
       add(b, a, c);
   }

   int l = 0, r = 1e9;
   while (l < r)
   {
       int mid = l + r >> 1;
       if (check(mid)) r = mid;
       else l = mid + 1;
   }

   printf("%d\n", l);
   return 0;
}

2. 聪明的质检员

来源：NOIP2011提高组 https://ac.nowcoder.com/acm/contest/259/B

算法知识点：二分，前缀和

复杂度： O(n^2)

#include 
#include 
#include 
#include 
using namespace std;
 
typedef long long LL; const int N = 200010;
 
int n, m;
LL S;
int w[N], v[N];
int l[N], r[N];
int cnt[N];
LL sum[N];
 
LL get(int W)
{
    for (int i = 1; i <= n; i++)
        if (w[i] >= W)
        {
            sum[i] = sum[i - 1] + v[i];
            cnt[i] = cnt[i - 1] + 1;
        }
    else
    {
        sum[i] = sum[i - 1];
        cnt[i] = cnt[i - 1];
    }
 
    LL res = 0;
    for (int i = 0; i < m; i++) res += (cnt[r[i]] - cnt[l[i] - 1]) *(sum[r[i]] - sum[l[i] - 1]);
    return res;
}
 
int main()
{
    scanf("%d%d%lld", &n, &m, &S);
    for (int i = 1; i <= n; i++) scanf("%d%d", &w[i], &v[i]);
    for (int i = 0; i < m; i++) scanf("%d%d", &l[i], &r[i]);
 
    int l = 0, r = 1e6 + 1;
    while (l < r)
    {
        int mid = l + r + 1 >> 1;
        if (get(mid) >= S) l = mid;
        else r = mid - 1;
    }
 
    printf("%lld\n", min(abs(get(r) - S), abs(S - get(r + 1))));
 
    return 0;
}

3.借教室

来源：NOIP2012提高组 https://ac.nowcoder.com/acm/contest/260/B

算法知识点：二分，差分

复杂度： O((n + m)logm

解题思路：

C++代码

#include 
#include 
#include 
#include 
using namespace std;
 
typedef long long LL; const int N = 1000010;
 
int n, m;
int r[N], d[N], s[N], t[N];
LL b[N];
 
bool check(int k)
{
    for (int i = 1; i <= n; i++) b[i] = r[i];
 
    for (int i = 1; i <= k; i++)
    {
        b[s[i]] -= d[i];
        b[t[i] + 1] += d[i];
    }
 
    LL res = 0;
    for (int i = 1; i <= n; i++)
    {
        res += b[i];
        if (res < 0) return true;
    }
 
    return false;
}
 
int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d", &r[i]);
    for (int i = n; i; i--) r[i] -= r[i - 1];
 
    for (int i = 1; i <= m; i++) scanf("%d%d%d", &d[i], &s[i], &t[i]);
 
    int l = 1, r = m;
    while (l < r)
    {
        int mid = l + r >> 1;
        if (check(mid)) r = mid;
        else l = mid + 1;
    }
 
    if (check(r))
    {
        puts("-1");
        printf("%d\n", r);
    }
    else puts("0");
 
    return 0;
}

4. 跳石头

来源：NOIP2015提高组 https://ac.nowcoder.com/acm/contest/263/A

算法知识点：二分，贪心

复杂度：O(NlogL)

解题思路：

C++代码

#include 
#include 
#include 
#include 
using namespace std; const int N = 50010;
 
int L, n, m;
int d[N];
 
bool check(int mid)
{
    int last = 0, cnt = 0;
    for (int i = 1; i <= n; i++)
        if (d[i] - last < mid) cnt++;
        else last = d[i];
    return cnt <= m;
}
 
int main()
{
    scanf("%d%d%d", &L, &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d", &d[i]);
    d[++n] = L;
 
    int l = 1, r = 1e9;
    while (l < r)
    {
        int mid = l + r + 1 >> 1;
        if (check(mid)) l = mid;
        else r = mid - 1;
    }
 
    printf("%d\n", r);
    return 0;
}

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