nowcoder牛客wannafly挑战赛20

A---染色

签到题，设最终颜色为x，一次操作就需要把一个不是x的点变为x，所以最终颜色为x时需要操作 总结点个数-颜色为x的节点个数，然后枚举所有颜色就行了

#include <iostream>
#include <string.h>
#include <cstdio>
#include <vector>
#include <map>
#include <math.h>
#include <string>
#include <algorithm>
#include <time.h>
 
#define SIGMA_SIZE 26
#define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) (x&-x)
#define foe(i, a, b) for(int i=a; i<=b; i++)
#define fo(i, a, b) for(int i = a; i < b; i++);
#pragma warning ( disable : 4996 )
 
using namespace std;
typedef long long LL;
inline LL LMax(LL a,LL b)      { return a>b?a:b; }
inline LL LMin(LL a,LL b)      { return a>b?b:a; }
inline LL lgcd( LL a, LL b ) { return b==0?a:lgcd(b,a%b); }
inline LL llcm( LL a, LL b ) { return a/lgcd(a,b)*b; }  //a*b = gcd*lcm
inline int Max(int a,int b)    { return a>b?a:b; }
inline int Min(int a,int b)    { return a>b?b:a; }
inline int gcd( int a, int b ) { return b==0?a:gcd(b,a%b); }
inline int lcm( int a, int b ) { return a/gcd(a,b)*b; }  //a*b = gcd*lcm
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL mod  = 1000000007;
const double eps = 1e-8;
const int inf  = 0x3f3f3f3f;
const int maxk = 1e6+5;
const int maxn = 1e5+5;
 
int N, ct;
int cnt[maxn];
LL num[maxn];
LL all;
struct col {
    int id;
    LL val;
    col() {}
}color[maxn];
 
bool cmp(const col& a, const col& b)
{
    return a.val < b.val;
}
 
void read()
{
    memset(cnt, 0, sizeof(cnt));
    memset(num, 0, sizeof(num));
    all = 0;
 
    foe(i, 1, N)
    {
        scanf("%lld", &color[i].val);
        all += color[i].val;
        color[i].id = i;
    }
    sort(color+1, color+1+N, cmp);
 
    int x, y;
    foe(i, 1, N-1)
    {
        scanf("%d %d", &x, &y);
    }
}
 
int main()
{
    while (~scanf("%d", &N))
    {
        read();
 
        LL tmp = color[1].val;
        num[1] = color[1].val;
        cnt[1] = 1;
        ct = 1;
        foe(i, 2, N)
        {
            if (color[i].val == tmp)
            {
                cnt[ct]++;
                continue;
            }
             
            tmp = color[i].val;
            num[++ct] = tmp;
            cnt[ct] = 1;
        }
 
        LL mmin = INF;
        foe(i, 1, ct)
        {
            tmp = all;
            tmp -= num[i]*cnt[i];
            if (tmp >= mmin) continue;
            tmp += (N-cnt[i])*num[i];
            mmin = LMin(tmp, mmin);
        }
        printf("%lld
", mmin);
    }
    return 0;
}

B---背包

我觉得这道题有点问题...先按价值排个序，然后考虑中位数两边最小的体积是多少呢，用优先队列从1~N扫一遍，再从N~1扫一遍，就可以记录下每个点往左和往右的所有物品中，取体积最小的M/2个物品的体积总和是多少，分别记为sum1[i],sum2[i]，然后分奇偶讨论一下就行了（还是觉得数据有点问题，）

#include <iostream>
#include <string.h>
#include <cstdio>
#include <vector>
#include <map>
#include <queue>
#include <math.h>
#include <string>
#include <algorithm>
#include <time.h>

#define SIGMA_SIZE 26
#define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) (x&-x)
#define foe(i, a, b) for(int i=a; i<=b; i++)
#define fo(i, a, b) for(int i = a; i < b; i++);
#pragma warning ( disable : 4996 )

using namespace std;
typedef long long LL;
inline LL LMax(LL a, LL b) { return a>b ? a : b; }
inline LL LMin(LL a, LL b) { return a>b ? b : a; }
inline LL lgcd(LL a, LL b) { return b == 0 ? a : lgcd(b, a%b); }
inline LL llcm(LL a, LL b) { return a / lgcd(a, b)*b; }  //a*b = gcd*lcm
inline int Max(int a, int b) { return a>b ? a : b; }
inline int Min(int a, int b) { return a>b ? b : a; }
inline int gcd(int a, int b) { return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b) { return a / gcd(a, b)*b; }  //a*b = gcd*lcm
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL mod = 1000000007;
const double eps = 1e-8;
const int inf = 0x3f3f3f3f;
const int maxk = 1e6 + 5;
const int maxn = 1e5 + 5;

//优先队列默认从大到排列
priority_queue<LL> qq;
int V, N, M;
LL sum;
LL sum1[maxn], sum2[maxn];
struct node {
    LL val, cost;
}pp[maxn];

bool cmp(const node& a, const node& b)
{
    return a.val == b.val ? a.cost<b.cost : a.val<b.val;
}

void init()
{
    foe(i, 1, N)
        scanf("%lld %lld", &pp[i].val, &pp[i].cost);
    sort(pp + 1, pp + 1 + N, cmp);
}

void solve(int i, int mid, LL ss[])
{
    ss[i] = sum;

    sum += pp[i].cost;
    qq.push(pp[i].cost);

    if (qq.size() > mid) {
        sum -= qq.top();
        qq.pop();
    }
}

int main()
{
    cin >> V >> N >> M;

    init();

    sum = 0;
    int mid = M / 2;
    //正向扫一遍
    for (int i = 1; i <= N; i++)
        solve(i, mid, sum1);

    sum = 0; while (!qq.empty()) qq.pop();
    //反向扫一遍
    for (int i = N; i >= 1; i--)
        solve(i, mid, sum2);


    if (M % 2) 
    {
        for (int i = N - mid; i >= mid + 1; i--)
            if (sum1[i] + sum2[i] + pp[i].cost <= V)
            {
                printf("%lld
", pp[i].val);
                break;
            }
    }
    else 
    {
        for (int i = N - mid + 1; i >= mid; i--)
            if (sum1[i] + sum2[i - 1] <= V)
            {
                printf("%lld
", (pp[i].val + pp[i - 1].val) / 2);
                break;
            }

    }


    return 0;
}