NOIP2010 题解

 1 #include <cstdio>
 2 #include <cstring>
 3 
 4 const int MAXN = 1000;
 5 
 6 int n, m, cj, now, MinV, MinId, Pri, a[MAXN+10], in[MAXN+10];
 7 
 8 int main(){
 9     memset(in, -1, sizeof(in));
10     scanf("%d %d", &n, &m), now = Pri = 0;
11     for (int i=0; i<m; i++){
12         scanf("%d", &cj);
13         if (~in[cj]) continue;
14         Pri ++;
15         if (now+1<=n) now ++, in[cj] = i;
16         else {
17             MinV = 0x3f3f3f3f;
18             for (int j=0; j<=MAXN; j++)
19                 if (in[j]<MinV && in[j]!=-1) MinV = in[j], MinId = j;
20             in[MinId] = -1, in[cj] = i; 
21         } 
22     }
23     printf("%d\n", Pri);
24 }

 1 #include <cstdio>
 2 #include <cstring>
 3 
 4 const int MAXN = 350+2;
 5 const int MAXE = 40+2;
 6 
 7 int n, m, cj, a[MAXN], b[6], dp[MAXE][MAXE][MAXE][MAXE];
 8 
 9 int main(){
10     memset(dp, 0, sizeof(dp));
11     
12     scanf("%d %d", &n, &m);
13     for (register int i=0; i<n; i++)
14         scanf("%d", &a[i]);
15     for (register int i=0; i<m; i++)
16         scanf("%d", &cj), b[cj] ++;
17 
18     dp[0][0][0][0] = a[0];
19     for (register int A=0; A<=b[1]; A++)
20         for (register int B=0; B<=b[2]; B++)
21             for (register int C=0; C<=b[3]; C++)
22                 for (register int D=0; D<=b[4]; D++){
23                     cj = A+B+B+C+C+C+D+D+D+D;
24                     if (A && dp[A-1][B][C][D]+a[cj]>dp[A][B][C][D])    dp[A][B][C][D] = dp[A-1][B][C][D]+a[cj];
25                     if (B && dp[A][B-1][C][D]+a[cj]>dp[A][B][C][D]) dp[A][B][C][D] = dp[A][B-1][C][D]+a[cj];
26                     if (C && dp[A][B][C-1][D]+a[cj]>dp[A][B][C][D]) dp[A][B][C][D] = dp[A][B][C-1][D]+a[cj];
27                     if (D && dp[A][B][C][D-1]+a[cj]>dp[A][B][C][D]) dp[A][B][C][D] = dp[A][B][C][D-1]+a[cj];
28                 }
29     printf("%d\n", dp[b[1]][b[2]][b[3]][b[4]]);
30 }

 1 #include <cstdio>
 2 #include <algorithm>
 3 using std::sort;
 4 
 5 const int MAXN = 20000+10;
 6 const int MAXM = 100000+10;
 7 
 8 int n, m, f[MAXN*2];
 9 
10 struct Relation{
11     int a, b, c;
12 
13     friend bool operator < (const Relation& A, const Relation& B){
14         return A.c>B.c;
15     } 
16 }r[MAXM];
17 
18 inline int Find(int x){
19     return f[x]==x ? x : f[x] = Find(f[x]);
20 }
21 
22 inline int Solve(){
23     sort(r, r+m);
24     int a, b;
25     for (int i=0; i<m; i++){
26         a = Find(r[i].a), b = Find(r[i].b);
27         if (a==b) return r[i].c;
28         else f[a] = Find(r[i].b+n), f[b] = Find(r[i].a+n);
29     }
30     return 0;
31 }
32 
33 int main(){
34     scanf("%d %d", &n, &m);
35     for (int i=1; i<=2*n; i++)
36         f[i] = i;
37     for (int i=0; i<m; i++)
38         scanf("%d %d %d", &r[i].a, &r[i].b, &r[i].c);
39 
40     printf("%d\n", Solve());
41 }