CSU-ACM2014年校队选拔赛指导赛解题报告

•Problem A  CSU 1065                               贪心

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 using namespace std;
 5 const int maxn = 1000010;
 6 struct Node{
 7     int a,b;
 8     bool operator < (const Node& rhs) const{
 9         return b < rhs.b;
10     }
11 }node[maxn];
12
13 int main(){
14     int n,r = -1,ans = 0;
15     scanf("%d",&n);
16     for(int i = 0;i < n;i++){
17         scanf("%d%d",&node[i].a,&node[i].b);
18     }
19     sort(node,node+n);
20     for(int i = 0;i < n;i++){
21         if(node[i].a > r){
22             ans++;
23             r = node[i].b;
24         }
25     }
26     printf("%d\n",ans);
27     return 0;
28 }

•Problem B  CSU 1060                                DP

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 using namespace std;
 5 const int maxn = 110;
 6 char s1[maxn],s2[maxn],s3[maxn];
 7 int dp[maxn][maxn][maxn];
 8
 9 int main(){
10     while(scanf("%s",s1+1) != EOF){
11         scanf("%s",s2+1);
12         scanf("%s",s3+1);
13         int len1 = strlen(s1+1);
14         int len2 = strlen(s2+1);
15         int len3 = strlen(s3+1);
16         memset(dp,0,sizeof(dp));
17
18         for(int i = 1;i <= len1;i++){
19             for(int j = 1;j <= len2;j++){
20                 for(int k = 1;k <= len3;k++){
21                     if(s1[i] == s2[j] && s2[j] == s3[k]){
22                         dp[i][j][k] = dp[i-1][j-1][k-1]+1;
23                     }else{
24                         int tmp1 = dp[i-1][j][k];
25                         int tmp2 = dp[i][j-1][k];
26                         int tmp3 = dp[i][j][k-1];
27                         tmp1 = max(tmp1,tmp2);
28                         tmp1 = max(tmp1,tmp3);
29                         dp[i][j][k] = tmp1;
30                     }
31                 }
32             }
33         }
34
35         printf("%d\n",dp[len1][len2][len3]);
36     }
37     return 0;
38 }

•Problem C  CSU 1307                               最短路

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <queue>
 5 using namespace std;
 6 typedef pair<int,int> pii;
 7 const int maxn = 2010;
 8 const int maxm = 100010;
 9 int v[maxm],next[maxm],w[maxm];
10 int d[maxm],first[maxn],e;
11 void init(){
12     e = 0;
13     memset(first,-1,sizeof(first));
14 }
15
16 void add_edge(int a,int b,int c){
17     v[e] = b;next[e] = first[a];w[e] = c;first[a] = e++;
18 }
19
20 int dijkstra(int src,int dist,int mid){
21     priority_queue <pii,vector<pii>,greater<pii> > q;
22     memset(d,-1,sizeof(d));
23     d[src] = 0;
24     q.push(make_pair(0,src));
25     while(!q.empty()){
26         while(!q.empty() && q.top().first > d[q.top().second])  q.pop();
27         if(q.empty())   break;
28         int u = q.top().second;
29         q.pop();
30         for(int i = first[u];i != -1;i = next[i])if(w[i] <= mid){
31             if(d[v[i]] > d[u] + w[i] || d[v[i]] == -1){
32                 d[v[i]] = d[u]+w[i];
33                 q.push(make_pair(d[v[i]],v[i]));
34             }
35         }
36     }
37     return d[dist];
38 }
39
40 int main(){
41     int n,m,src,dist;
42     while(scanf("%d%d%d%d",&n,&m,&src,&dist) != EOF){
43         init();
44         int l = 1<<30,r = -1,ans = 1<<30;
45         for(int i = 0;i < m;i++){
46             int a,b,c;
47             scanf("%d%d%d",&a,&b,&c);
48             add_edge(a,b,c);
49             add_edge(b,a,c);
50             l = min(l,c);
51             r = max(r,c);
52         }
53         while(l <= r){
54             int mid = (l+r)>>1;
55             int tmp = dijkstra(src,dist,mid);
56             if(tmp == -1)    l = mid+1;
57             else    r = mid-1,ans = tmp;
58         }
59
60         printf("%d\n",ans);
61     }
62     return 0;
63 }

•Problem D  CSU 1290                               DP

 1 #include <cstdio>
 2 double dp[1005];
 3 int main()
 4 {
 5     int n,t;
 6     double k;
 7     scanf("%d", &t);
 8     while (t--) {
 9         scanf("%lf %d", &k, &n);
10         dp[1] = 1.0;
11         for (int i = 2; i <= n; ++i)
12             dp[i] = dp[i-1] + (k - dp[i-1])/k;
13         printf("%.5f\n", dp[n]);
14     }
15     return 0;
16 }

CSU-ACM2014年校队选拔赛指导赛解题报告