Problem A Football
题意：给n个球队，每个球队之间只能有一场比赛。要求构造一个每个球队只胜k场。
思路：直接暴力，标记一下就可以了。
代码如下：
1 /************************************************** 2 * Author : xiaohao Z 3 * Blog : http://www.cnblogs.com/shu-xiaohao/ 4 * Last modified : 2014-04-17 23:25 5 * Filename : RCC_2014_W_A.cpp 6 * Description : 7 * ************************************************/ 8 9 #include <iostream> 10 #include <cstdio> 11 #include <cstring> 12 #include <cstdlib> 13 #include <cmath> 14 #include <algorithm> 15 #include <queue> 16 #include <stack> 17 #include <vector> 18 #include <set> 19 #include <map> 20 #define MP(a, b) make_pair(a, b) 21 #define PB(a) push_back(a) 22 23 using namespace std; 24 typedef long long ll; 25 typedef pair<int, int> pii; 26 typedef pair<unsigned int,unsigned int> puu; 27 typedef pair<int, double> pid; 28 typedef pair<ll, int> pli; 29 typedef pair<int, ll> pil; 30 31 const int INF = 0x3f3f3f3f; 32 const double eps = 1E-6; 33 const int LEN = 1010; 34 int Map[LEN][LEN]; 35 36 int main() 37 { 38 //freopen("in.txt", "r", stdin); 39 40 int n, m; 41 while(scanf("%d%d", &n, &m)!=EOF){ 42 memset(Map, 0, sizeof Map); 43 int ans = 1, anscnt = 0; 44 for(int i=0; i<n; i++){ 45 int cnt = 0; 46 for(int j=0; j<n; j++){ 47 if(Map[i][j] != 0 || i==j) continue; 48 if(cnt == m) break; 49 Map[i][j] = 1; 50 Map[j][i] = -1; 51 cnt++;anscnt ++; 52 } 53 if(cnt != m) { 54 ans = 0; 55 break; 56 } 57 } 58 if(!ans) printf("-1 "); 59 else { 60 printf("%d ", anscnt); 61 for(int i=0; i<n; i++) 62 for(int j=0; j<n; j++) 63 if(Map[i][j] == 1) 64 printf("%d %d ", i+1, j+1); 65 } 66 67 } 68 return 0; 69 }
Problem B Cunning Gena
思路：状压DP dp[i]表示状态为i时的最小花费。先对朋友要求显示器数量排序。然后开始DP（和背包比较像）
代码如下：
1 /************************************************** 2 * Author : xiaohao Z 3 * Blog : http://www.cnblogs.com/shu-xiaohao/ 4 * Last modified : 2014-04-17 23:25 5 * Filename : RCC_2014_W_B.cpp 6 * Description : 7 * ************************************************/ 8 9 #include <iostream> 10 #include <cstdio> 11 #include <cstring> 12 #include <cstdlib> 13 #include <cmath> 14 #include <algorithm> 15 #include <queue> 16 #include <stack> 17 #include <vector> 18 #include <set> 19 #include <map> 20 #define MP(a, b) make_pair(a, b) 21 #define PB(a) push_back(a) 22 23 using namespace std; 24 typedef long long ll; 25 typedef pair<int, int> pii; 26 typedef pair<unsigned int,unsigned int> puu; 27 typedef pair<int, double> pid; 28 typedef pair<ll, int> pli; 29 typedef pair<int, ll> pil; 30 31 const int INF = 0x3f3f3f3f; 32 const ll LINF = 0x3f3f3f3f3f3f3f3fLL; 33 const double eps = 1E-6; 34 const int LEN = 21; 35 ll dp[1<<21]; 36 struct F{ 37 int cost, mon, st; 38 }f[101]; 39 40 bool cmp(F a, F b){ 41 return a.mon < b.mon; 42 } 43 44 int main() 45 { 46 // freopen("in.txt", "r", stdin); 47 48 int n, m, b, tn, tc, tm, tmp; 49 while(scanf("%d%d%d", &n, &m, &b)!=EOF){ 50 for(int i=0; i<n; i++){ 51 scanf("%d%d%d", &f[i].cost, &f[i].mon, &tn); 52 f[i].st = 0; 53 for(int j=0; j<tn; j++){ 54 scanf("%d", &tmp); 55 f[i].st |= (1 << (tmp-1)); 56 } 57 } 58 sort(f, f+n, cmp); 59 for(int i=0; i<(1<<21); i++) dp[i] = LINF; 60 dp[0] = 0; 61 ll ans = LINF; 62 for(int i=0; i<n; i++){ 63 for(int j=(1 << m)-1; j>=0; j--){ 64 dp[j|f[i].st] = min(dp[j|f[i].st], dp[j] + f[i].cost); 65 } 66 ans = min(dp[(1<<m)-1]+(ll)f[i].mon*(ll)b, ans); 67 } 68 if(ans == LINF) printf("-1 "); 69 else printf("%I64d ", ans); 70 } 71 return 0; 72 }