2014上海邀请赛

Problem A. Game with pearls http://acmoj.shu.edu.cn/openjudge/viewproblem.php?coll_id=1&prob_id=350

贪心，每次选最小，看能否满足，用优先队列维护。

 1 #include<cstdio>
 2 #include<queue>
 3 using namespace std;
 4 struct G{
 5     int val;
 6     friend bool operator <(G a,G b){
 7         return a.val>b.val;
 8     }
 9 }now;
10 priority_queue<G> q;
11 int main(){
12     int t,n,k,x;
13     while(~scanf("%d",&t)){
14         while(t--){
15             scanf("%d%d",&n,&k);
16             while(!q.empty()) q.pop();
17             for(int i=0;i<n;i++){
18                 scanf("%d",&x);
19                 now.val=x;
20                 q.push(now);
21             }
22             bool ans=true;
23             for(int i=1;i<=n;i++){
24                 bool flag=false;
25                 while(!q.empty()){
26                     now=q.top();
27                     q.pop();
28                     if(now.val>i){
29                         break;
30                     }
31                     while(true){
32                         if(now.val>=i) break;
33                         now.val+=k;
34                     }
35                     if(now.val==i){
36                         flag=true;
37                         break;
38                     }
39                     q.push(now);
40                 }
41                 if(!flag){
42                     ans=false;
43                     break;
44                 }
45             }
46             if(ans) puts("Jerry");
47             else puts("Tom");
48         }
49     }
50     return 0;
51 }

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Problem C. Seam Carving http://acmoj.shu.edu.cn/openjudge/viewproblem.php?coll_id=1&prob_id=352

dp求二维数组一列和最小，并记录路径，优先选右侧的点，dp[i][j]=max(dp[i-1][j+1],dp[i-1][j],dp[i-1][j-1])+a[i][j];

 1 #include<cstdio>
 2 #include<cstring>
 3 #define mt(a,b) memset(a,b,sizeof(a))
 4 const int inf=0x3f3f3f3f;
 5 const int M=128;
 6 int a[M][M];
 7 int dp[M][M];
 8 int pre[M][M];
 9 int dy[]={1,0,-1};
10 int ans[M];
11 int main(){
12     int t,n,m;
13     while(~scanf("%d",&t)){
14         int cas=1;
15         while(t--){
16             scanf("%d%d",&n,&m);
17             mt(dp,0);
18             for(int i=1;i<=n;i++){
19                 for(int j=1;j<=m;j++){
20                     scanf("%d",&a[i][j]);
21                     dp[i][j]=inf;
22                 }
23             }
24             mt(pre,-1);
25             for(int i=1;i<=n;i++){
26                 for(int j=1;j<=m;j++){
27                     for(int k=0;k<3;k++){
28                         int ty=j+dy[k];
29                         if(ty>=1&&ty<=m){
30                             if(dp[i][j]>dp[i-1][ty]+a[i][j]){
31                                 dp[i][j]=dp[i-1][ty]+a[i][j];
32                                 pre[i][j]=k;
33                             }
34                         }
35                     }
36                 }
37             }
38             int sma=inf;
39             int id=-1;
40             for(int i=1;i<=m;i++){
41                 if(sma>=dp[n][i]){
42                     sma=dp[n][i];
43                     id=i;
44                 }
45             }
46             int nx=n,ny=id,la=0;
47             while(~pre[nx][ny]){
48                 ans[la++]=ny;
49                 int k=pre[nx][ny];
50                 ny+=dy[k];
51                 nx--;
52             }
53             printf("Case %d\n",cas++);
54             for(int i=la-1;i>=0;i--){
55                 if(i!=la-1) printf(" ");
56                 printf("%d",ans[i]);
57             }
58             puts("");
59         }
60     }
61     return 0;
62 }

View Code

Problem D. Battle ships http://acmoj.shu.edu.cn/openjudge/viewproblem.php?coll_id=1&prob_id=353

竟然是二分图匹配，巧妙的标号和匹配。

  1 #include<cstdio>
  2 #include<cstring>
  3 #define mt(a,b) memset(a,b,sizeof(a))
  4 const int M=3000;
  5 char a[M][M];
  6 int heng[M][M];
  7 int shu[M][M];
  8 class Bipartite_graph {  //二分图最大匹配  匈牙利算法
  9     bool mat[M][M],vis[M];
 10     int res[M],n,m,big;
 11     bool dfs(int a) {
 12         for (int i=1; i<=m; i++) {
 13             if (mat[a][i]&&!vis[i]) {
 14                 vis[i]=true;
 15                 if(!res[i]||dfs(res[i])) {
 16                     res[i]=a;
 17                     return true;
 18                 }
 19             }
 20         }
 21         return false;
 22     }
 23 public:
 24     void init(int x,int y) {
 25         n=x;
 26         m=y;
 27         mt(mat,0);
 28         mt(res,0);
 29     }
 30     void add(int u,int v) {
 31         mat[u][v]=true;
 32     }
 33     int solve() {
 34         big=0;
 35         for(int i=1; i<=n; i++) {
 36             mt(vis,0);
 37             if(dfs(i)) big++;
 38         }
 39         return big;
 40     }
 41 } gx;
 42 int main(){
 43     int t,n,m;
 44     while(~scanf("%d",&t)){
 45         while(t--){
 46             scanf("%d%d",&m,&n);
 47             for(int i=0;i<m;i++){
 48                 scanf("%s",a[i]);
 49             }
 50             mt(heng,0);
 51             mt(shu,0);
 52             int cnt=1;
 53             for(int i=0;i<m;i++){
 54                 for(int j=0;j<n;j++){
 55                     if(a[i][j]=='*'){
 56                         heng[i][j]=cnt;
 57                     }
 58                     else if(a[i][j]=='#'){
 59                         cnt++;
 60                     }
 61                 }
 62                 cnt++;
 63             }
 64             int pre=-1;
 65             cnt=0;
 66             for(int i=0;i<m;i++){
 67                 for(int j=0;j<n;j++){
 68                     if(heng[i][j]){
 69                         if(heng[i][j]==pre){
 70                             heng[i][j]=cnt;
 71                         }
 72                         else{
 73                             cnt++;
 74                             pre=heng[i][j];
 75                             heng[i][j]=cnt;
 76                         }
 77                     }
 78                 }
 79             }
 80             int nn=cnt;
 81             cnt=1;
 82             for(int j=0;j<n;j++){
 83                 for(int i=0;i<m;i++){
 84                     if(a[i][j]=='*'){
 85                         shu[i][j]=cnt;
 86                     }
 87                     else if(a[i][j]=='#'){
 88                         cnt++;
 89                     }
 90                 }
 91                 cnt++;
 92             }
 93             pre=-1,cnt=0;
 94             for(int j=0;j<n;j++){
 95                 for(int i=0;i<m;i++){
 96                     if(shu[i][j]){
 97                         if(shu[i][j]==pre){
 98                             shu[i][j]=cnt;
 99                         }
100                         else{
101                             cnt++;
102                             pre=shu[i][j];
103                             shu[i][j]=cnt;
104                         }
105                     }
106                 }
107             }
108             gx.init(nn,cnt);
109             for(int i=0;i<m;i++){
110                 for(int j=0;j<n;j++){
111                     if(heng[i][j]&&shu[i][j]){
112                         gx.add(heng[i][j],shu[i][j]);
113                     }
114                 }
115             }
116             printf("%d\n",gx.solve());
117         }
118     }
119     return 0;
120 }

View Code

Problem F. Linearization of the kernel functions in SVM http://acmoj.shu.edu.cn/openjudge/viewproblem.php?coll_id=1&prob_id=355

模拟输出，无算法，有细节。

 1 #include<cstdio>
 2 int a[16];
 3 char op[]={"pqruvwxyz"};
 4 int main(){
 5     int n;
 6     while(~scanf("%d",&n)){
 7         while(n--){
 8             for(int i=0;i<10;i++){
 9                 scanf("%d",&a[i]);
10             }
11             bool flag=false;
12             for(int i=0;i<9;i++){
13                 if(a[i]){
14                     if(flag&&a[i]>0) printf("+");
15                     if(a[i]==-1) printf("-");
16                     if(a[i]!=-1&&a[i]!=1) printf("%d",a[i]);
17                     putchar(op[i]);
18                     flag=true;
19                 }
20             }
21             if(a[9]){
22                 if(flag&&a[9]>0) printf("+");
23                 printf("%d",a[9]);
24                 flag=true;
25             }
26             if(!flag) printf("0");
27             puts("");
28         }
29     }
30     return 0;
31 }

View Code

Problem J. Comparison of Android versions http://acmoj.shu.edu.cn/openjudge/viewproblem.php?coll_id=1&prob_id=359

模拟比较，无算法，无细节。

 1 #include<cstdio>
 2 #include<cstring>
 3 const int M=16;
 4 char a[M],b[M],cma[M],cmb[M];
 5 char op[]={"<=>"};
 6 int main(){
 7     int n;
 8     while(~scanf("%d",&n)){
 9         for(int i=1;i<=n;i++){
10             scanf("%s%s",a,b);
11             printf("Case %d: ",i);
12             cma[0]=a[0];
13             cmb[0]=b[0];
14             cma[1]=0;
15             cmb[1]=0;
16             printf("%c ",op[1+strcmp(cma,cmb)]);
17             int la=0,lb=0;
18             for(int j=2;j<5;j++){
19                 cma[la++]=a[j];
20                 cmb[lb++]=b[j];
21             }
22             if(a[1]==b[1]){
23                 cma[la++]=a[5];
24                 cmb[lb++]=b[5];
25             }
26             cma[la]=0;
27             cmb[lb]=0;
28             printf("%c\n",op[1+strcmp(cma,cmb)]);
29         }
30     }
31     return 0;
32 }

View Code

posted on 2014-07-25 10:03 gaolzzxin 阅读(196) 评论(0) 编辑收藏举报