UESTC 2016 Summer Training #18 Div.2（未完待续）

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题目来源：http://codeforces.com/gym/100257
2013-2014 ACM-ICPC, NEERC, Moscow Subregional Contest

A
题意：给你N个点，以及每个点的钥匙数，从A点走到B点需要消耗1把A型钥匙，问你最多能走出多少个三角形。
解法：用优先队列维护，优先队列中放的是所有的非零钥匙数，一个三角形只需要（1，1，1）或者（0，1，2）这两种之一就可以形成。
先对（0，1，2）进行处理，每次取出最大的两个数，更新答案和这两个数，知道最大的数小于2为止。
然后对（1，1，1）进行类似处理，每次取出三个1，更新答案并且把这三个1变为0。
这两部分算完后就是答案。

/* UESTC Summer Training #18 Div.2. Problem A. */#include <bits/stdc++.h>#include <cstdio>#include <queue>#define MIN(a, b) ( ((a)<(b))?(a):(b) )#define MAXN  10024using namespace std;int N = 0;int nfields = 0;priority_queue<int> keys;int main(void){    int i = 0;    int t = 0;    int k[3] = {0};    scanf("%d", &N);    while(!keys.empty())        keys.pop();    for(i = 0;i < N;++i)    {        scanf("%d", &t);        keys.push(t);    }    if(N < 3)    {        puts("0");    }    else    {        while(1)        {            k[0] = keys.top();            if(k[0] < 2)                break;            if(keys.size() < 2)                break;            for(i = 0;i < 2;++i)            {                k[i] = keys.top();                keys.pop();            }            int s = min(k[1],k[0] / 2);            nfields += s;            k[0] -= 2 * s;            k[1] -= s;            //printf("k0 = %d k1 = %d\n",k[0],k[1]);            if(k[0] != 0)                keys.push(k[0]);            if(k[1] != 0)                keys.push(k[1]);            //printf(" %d\n", nfields);        }        //printf("%d\n",keys.size());        while( keys.size() > 2 )        {            for(i = 0;i < 3;++i)            {                k[i] = keys.top();                keys.pop();            }            nfields++;            for(i = 0;i < 3;++i)            {                if( (--k[i]) > 0 )                    keys.push(k[i]);            }            //printf("  %d\n", nfields);        }        printf("%d\n", nfields);    }    return 0;}

B
题意：给一个16进制数，让你找出16进制数中比这个数大且尽可能小且每一位都不相同的数是多少。

/* UESTC Summer Training #18 Div.2. Problem B. */#include <stdio.h>#include <string.h>const char digit[16] = {'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};char origin[32] = {'0',0};int l = 0;int inc(int i){  int count = 0;  while( i >= 0 )  {    if( 'F' == origin[i] )    {      count++;      origin[i] = '0';      i--;    }    else    {      if( '9' == origin[i] )        origin[i] = 'A';      else        origin[i]++;      break;    }  }  return count;}void handle(int togenerate){  int i = 0;  int used[128] = {0};  int nused = 0;  for(i = (origin[0]>'0')?(0):(1);i < togenerate;++i)    if( 1 == used[origin[i]] )    {      handle(i+1-inc(i));      nused++;      break;    }    else      used[origin[i]] = 1;  if( 0 == nused )    if( togenerate < l )    {      int count = 0;      for(i = 0;i < 16 && count < (l-togenerate);++i)        if( 0 == used[digit[i]] )        {          origin[togenerate+count] = digit[i];          ++count;        }    }}int main(void){  scanf("%s", origin+1);  l = strlen(origin);  inc(l-1);  handle(l);  printf("%s\n", (origin[0]>'0')?(origin):(origin+1));  return 0;}

#include <cstdio>  #include <cstring>  #include <cstdlib>  #include <climits>  #include <string>  typedef unsigned long long ull;  typedef long long ll;  char s[111];  int n;  int go(int pos) {      if(pos==n) return 1;      int i;      for(i=0;i<pos;i++) if(s[i] == s[pos]) break;      if(i==pos && go(pos+1)) return 1;      for(char c=s[pos]+1;c!='G';c++) {          if(c==('9'+1)) c='A';          s[pos]=c;          for(i=0;i<pos;i++) if(s[i] == c) break;          if(i != pos) continue;          for(int j=pos+1;j<n;j++) s[j]='0';          if(go(pos+1)) return 1;      }      return 0;  }  int main() {      ull x;      while(scanf("%llx", &x) != EOF) {          x++;          sprintf(s, "%llx", x);          n = strlen(s);          for(int i=0;i<n;i++) s[i] = toupper(s[i]);          if(go(0)) printf("%s\n", s);          else {              memset(s, '0', sizeof(s));              s[0]='1';              n++;              go(0);              s[n]='\0';              printf("%s\n", s);          }      }      return 0;  }

#include <cstdio>  #include <cstring>  #include <algorithm>  #include <map>  #include <iostream>  #include <vector>  #include <queue>  #include <stack>  #include <set>  #include <string>  #define ll long long  using namespace std;  typedef pair<int, int> PII;  int a[2003], b[2003];  int dp[2003];  int n;  int main() {      int i, j;      scanf("%d", &n);      for (i = 1; i <= n; i++) {          int x, y;          char ch;          scanf("%d:%d %c", &x, &y, &ch);          a[i] = x * 60 + y;          b[i] = (ch == 'U');      }      int cnt = 0, last = -1e9;      bool flag = 0;      int pay = 0;      for (i = 1; i <= n; i++) {          if (a[i] - last <= 90) {              if (b[i] == 0) {                  if (!pay) {                      cnt += 26;                      pay = 26;                  } else {                      cnt += 44 - pay;                      pay = 44;                  }              } else if (!flag) {                  flag = 1;                  if (!pay) {                      cnt += 28;                      pay = 28;                  } else {                      cnt += 44 - pay;                      pay = 44;                  }              } else {                  last = a[i];                  cnt += 28;                  pay = 28;              }          } else {              last = a[i];              if (b[i] == 0) {                  cnt += 26;                  pay = 26;                  flag =0;              } else {                  flag = 1;                  cnt += 28;                  pay = 28;              }          }          //printf("cnt = %d ", cnt);      }      printf("%d ", cnt);      for (i = 0; i <= n; i++)          dp[i] = 1e9;      dp[0] = 0;      a[0] = -1e9;      for (i = 0; i < n; i++) {          dp[i + 1] = min(dp[i + 1], dp[i] + (b[i + 1] == 1 ? 28 : 26));      //  printf("%d////////\n", dp[i+1]);          int cnt = 0;          j = i + 1;          int cur = a[i + 1];          while (j <= n && a[j] - cur <= 90) {              if (b[j] == 1)                  cnt++;              if (cnt >= 2)                  break;              j++;          }          j--;          dp[j] = min(dp[j], dp[i] + 44);      }  //  for(i = 1; i <= n; i++)  //      printf("%d~~~ ", dp[i]);      printf("%d\n", dp[n]);      return 0;  }

#include <bits/stdc++.h>#define _ ios_base::sync_with_stdio(0);cin.tie(0);#define INF 0x3f3f3f3f#define eps 1e-6typedef long long LL;const double pi = acos(-1.0);const long long mod = 1e9 + 7;using namespace std;int fun(int a,int c,int b){    if(a >= b && b >= c || a <= b && b <= c)        return 1;    return 0;}int main(){    ios_base::sync_with_stdio(0);cin.tie(0);    int a[3],b[3];    cin >> a[0] >> b[0];    cin >> a[1] >> b[1];    cin >> a[2] >> b[2];    int ok = 0;    if(fun(a[0],a[1],a[2]) && fun(b[0],b[1],b[2]))        ok = 1;    if(ok)        puts("Yes");    else        puts("No");    return 0;}

I
题意：给你N个集合，并且给出这N个集合两两之间的交集，问你是否能构造出这样的N个集合。
解法：用set模拟即可。

#include <bits/stdc++.h>#define INF 0x3f3f3f3f#define eps 1e-6typedef long long LL;const double pi = acos(-1.0);const long long mod = 25 * 1E8;using namespace std;set <int> s[205];set <int> :: iterator it;int a[205][205];int main(){    //freopen("input.txt","r",stdin);    //freopen("output.txt","w",stdout);    ios_base::sync_with_stdio(0);cin.tie(0);    int N;    cin >> N;    memset(a,0,sizeof(a));    int x,y,k,p;    for(int i = 1;i <= (N * (N - 1) / 2);i++)    {        cin >> x >> y >> k;        a[x][y] = a[y][x] = k;        for(int i = 0;i < k;i++)        {            cin >> p;            s[x].insert(p);            s[y].insert(p);        }    }    int ok = 1;    for(int i = 1;i <= N;i++)        for(int j = i + 1;j <= N;j++){            int num = 0;            for(it = s[i].begin();it != s[i].end();it++)                if(s[j].find(*it) != s[j].end())                    num++;            if(num != a[i][j])            {                ok = 0;                break;            }    }    if(ok)    {        puts("Yes");        for(int i = 1;i <= N;i++)        {            int q = s[i].size();            printf("%d",q);            for(it = s[i].begin();it != s[i].end();it++)                printf(" %d",*it);            puts("");        }    }    else        puts("No");    return 0;}

#include <cstdio>#include <algorithm>int ans[200000], q[200000];int main() {    int a, b, c, x, y, n, k, l = 0;    scanf("%d%d%d%d%d%d%d", &n, &k, &x, &y, &a, &b, &c);    int tx = x, ty = y, tz;    for (int i = 0; i < n; i++) {        tz = a*tx+b*ty+c&(1<<31)-1;        tx = ty, ty = tz;        if (!(i%20)) {            if (l >= k && tz <= -q[0])                 continue;            if (l >= k)                 std::pop_heap(q, q+l), l--;            q[l++] = -tz;            std::push_heap(q, q+l);        }    }    tx = x, ty = y;    for (int i = 0; i < n; i++) {        tz = a*tx+b*ty+c&(1<<31)-1;        tx = ty, ty = tz;        if (i%20) {            if (l >= k && tz <= -q[0])                 continue;            if (l >= k)                 std::pop_heap(q, q+l), l--;            q[l++] = -tz;            std::push_heap(q, q+l);        }    }    for (int i = 0; i < k; i++)         ans[k-i-1] = -q[0], std::pop_heap(q, q+l), l--;    printf("%d", ans[0]);    for (int i = 1; i < k; i++)         printf(" %d", ans[i]);    puts("");    return 0;}

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