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==?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==?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  = ;
 const double eps = 1e-;
 const int inf  = 0x3f3f3f3f;
 const int maxk = 1e6+;
 const int maxn = 1e5+;

 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, , sizeof(cnt));
     memset(num, , sizeof(num));
     all = ;

     foe(i, , N)
     {
         scanf("%lld", &color[i].val);
         all += color[i].val;
         color[i].id = i;
     }
     sort(color+, color++N, cmp);

     int x, y;
     foe(i, , N-)
     {
         scanf("%d %d", &x, &y);
     }
 }

 int main()
 {
     while (~scanf("%d", &N))
     {
         read();

         LL tmp = color[].val;
         num[] = color[].val;
         cnt[] = ;
         ct = ;
         foe(i, , N)
         {
             if (color[i].val == tmp)
             {
                 cnt[ct]++;
                 continue;
             }

             tmp = color[i].val;
             num[++ct] = tmp;
             cnt[ct] = ;
         }

         LL mmin = INF;
         foe(i, , ct)
         {
             tmp = all;
             tmp -= num[i]*cnt[i];
             if (tmp >= mmin) continue;
             tmp += (N-cnt[i])*num[i];
             mmin = LMin(tmp, mmin);
         }
         printf("%lld\n", mmin);
     }
     return ;
 }

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 ==  ? 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 ==  ? 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 = ;
 const double eps = 1e-;
 const int inf = 0x3f3f3f3f;
 const int maxk = 1e6 + ;
 const int maxn = 1e5 + ;

 //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, , N)
         scanf("%lld %lld", &pp[i].val, &pp[i].cost);
     sort(pp + , pp +  + 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 = ;
     int mid = M / ;
     //正向扫一遍
     for (int i = ; i <= N; i++)
         solve(i, mid, sum1);

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

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

     }

     return ;
 }