Codeforces Round #336 (Div. 2)

水 A - Saitama Destroys Hotel

简单的模拟，小贪心。其实只要求max (ans, t + f);

#include <bits/stdc++.h>
using namespace std;

#define lson l, mid, o << 1
#define rson mid + 1, r, o << 1 | 1
typedef long long ll;
const int N = 1e5 + 5;
const int INF = 0x3f3f3f3f;

int n, s;
struct Floor    {
    int f, t;
    bool operator < (const Floor &r) const  {
        return f > r.f || (f == r.f && t < r.t);
    }
}a[105];

int main(void)  {
    scanf ("%d%d", &n, &s);
    for (int i=1; i<=n; ++i)    {
        scanf ("%d%d", &a[i].f, &a[i].t);
    }
    sort (a+1, a+1+n);
    int ans = 0, now = s, i = 1;
    while (now > 0 && i <= n) {
        if (now > a[i].f)   {
            now--;  ans++;
        }
        else    {
            if (ans < a[i].t)   {
                ans = a[i].t;   i++;
                while (i <= n && a[i].f == now && ans >= a[i].t)    i++;
            }
            else    {
                while (i <= n && a[i].f == now && ans >= a[i].t)    i++;
            }
        }
    }
    while (now > 0) {
        now--;  ans++;
    }
    printf ("%d
", ans);

    return 0;
}

组合 B - Hamming Distance Sum

题意：都是01的a字符串在b字符串的所有连续子串不同的个数。

分析：看上去很难做。但是换一个思路，考虑a字符串的每一个字符最终会和b中哪些字符比较，求一个前缀就能解决了。多想想还是能想出来的，JayYe说做多了这些题目都是同一个套路。

#include <bits/stdc++.h>
using namespace std;

#define lson l, mid, o << 1
#define rson mid + 1, r, o << 1 | 1
typedef long long ll;
const int N = 2e5 + 5;
const int INF = 0x3f3f3f3f;

char s[N], t[N];
int cnt0[N], cnt1[N];

int main(void)  {
    scanf ("%s%s", s + 1, t + 1);
    int lens = strlen (s + 1), lent = strlen (t + 1);
    memset (cnt0, 0, sizeof (cnt0));
    memset (cnt1, 0, sizeof (cnt1));
    int c1[2] = {0, 0}, c2[2] = {0, 0};
    for (int i=1; i<=lens; ++i)  if (s[i] == '0')    c1[0]++;
    for (int i=1; i<=lent; ++i)  {
        if (t[i] == '0')    {
            c2[0]++;    cnt0[i] = cnt0[i-1] + 1;
            cnt1[i] = cnt1[i-1];
        }
        else    {
            cnt1[i] = cnt1[i-1] + 1;
            cnt0[i] = cnt0[i-1];
        }
    }
    c1[1] = lens - c1[0];   c2[1] = lent - c2[0];
    ll ans = 0;
    for (int i=1; i<=lens; ++i)  {
        if (s[i] == '0')    {
            ans += cnt1[lent-lens+i] - cnt1[i-1];
        }
        else    {
            ans += cnt0[lent-lens+i] - cnt0[i-1];
        }
    }
    printf ("%I64d
", ans);

    return 0;
}

　　

DP C - Chain Reaction

题意：在最后面多一个becaon，问在某一个地方某一个能量能使最后死掉的becaon最少。

分析：假设会杀掉j的becaon，那么找出j位置会杀掉的位置，那么j到n都死完了，pos到j也死完了，再加上kill[pos]的求最小值，kill[]是dp，从后往前能更新。

#include <bits/stdc++.h>
using namespace std;

#define lson l, mid, o << 1
#define rson mid + 1, r, o << 1 | 1
typedef long long ll;
const int N = 1e5 + 5;
const int INF = 0x3f3f3f3f;

struct Beacon   {
    int a, b;
    bool operator < (const Beacon &r) const {
        return a < r.a;
    }
}p[N];
int kill[N];

int bsearch(int l, int r, int k)    {
    while (l <= r)  {
        int mid = l + r >> 1;
        if (p[mid].a < k)   l = mid + 1;
        else if (p[mid].a == k) return mid - 1;
        else    {
            r = mid - 1;
        }
    }
    return r;
}

int main(void)  {
    int n;  scanf ("%d", &n);
    for (int i=1; i<=n; ++i)    {
        scanf ("%d%d", &p[i].a, &p[i].b);
    }
    sort (p+1, p+1+n);
    int ans = n;
    for (int i=n-1; i>=0; --i)  {   //kill i bacon
        int j = n - i;
        if (j == 1) {
            ans = n - 1;    kill[1] = 0;
            continue;
        }
        int pos = bsearch (1, j - 1, p[j].a - p[j].b);
        while (pos > 0 && p[j].a - p[j].b <= p[pos].a)    pos--;
        if (pos == j - 1)   kill[j] = kill[j-1];
        else    kill[j] = kill[pos] + j - pos - 1;
        ans = min (ans, kill[pos] + j - pos - 1 + i);
    }
    printf ("%d
", ans);

    return 0;
}