2010 Asia Fuzhou Regional Contest

A hard Aoshu Problem http://acm.hdu.edu.cn/showproblem.php?pid=3699

用深搜写排列，除法要注意，还有不能有前导零。当然可以5个for，但是如果有很多个，dfs还是好的。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<map>
 5 #define mt(a,b) memset(a,b,sizeof(a))
 6 using namespace std;
 7 const int M=16;
 8 char sa[M],sb[M],sc[M],ha[M],hb[M],op[]="+-*/";
 9 int val[8];
10 bool use[16];
11 map<string,bool> mp;
12 string str;
13 bool prezero(char s[]){
14     if(strlen(s)>1&&!val[s[0]-'A']) return true; return false;
15 }
16 int getint(char s[]){
17     int res=0;
18     for(int i=0;s[i];i++){
19         res=(res<<3)+(res<<1)+val[s[i]-'A'];
20     }
21     return res;
22 }
23 int solve(const int &a,const char &ope,const int &b){
24     if(ope=='+') return a+b;
25     if(ope=='-') return a-b;
26     if(ope=='*') return a*b;
27     if(b&&!(a%b))return a/b;
28     return -1;
29 }
30 void flag(const int &a,const char &ope,const int &b){
31     sprintf(ha,"%d",a);
32     sprintf(hb,"%d",b);
33     str=(string)ha+ope+(string)hb;
34     mp[str]=true;
35 }
36 void dfs(int t){
37     if(t==5){
38         if(prezero(sa)||prezero(sb)||prezero(sc)) return ;
39         int a=getint(sa);
40         int b=getint(sb);
41         int c=getint(sc);
42         for(int i=0;i<4;i++){
43             if(solve(a,op[i],b)==c){
44                 flag(a,op[i],b);
45             }
46         }
47         return ;
48     }
49     for(int i=0;i<10;i++){
50         if(!use[i]){
51             val[t]=i;
52             use[i]=true;
53             dfs(t+1);
54             use[i]=false;
55         }
56     }
57 }
58 int main(){
59     int t;
60     while(~scanf("%d",&t)){
61         while(t--){
62             scanf("%s%s%s",sa,sb,sc);
63             mt(use,0);
64             mp.clear();
65             dfs(0);
66             printf("%d\n",mp.size());
67         }
68     }
69     return 0;
70 }

View Code

Fermat Point in Quadrangle http://acm.hdu.edu.cn/showproblem.php?pid=3694

全一个点跑不出来，特判一下。

 1 #include<cstdio>
 2 #include<cmath>
 3 const double eps=1e-8;
 4 struct point{
 5     double x,y;
 6 }p[8];
 7 class Fermatpoint {//多边形的费马点
 8 #define Q(x) ((x) * (x))
 9 #define EQU(x, y) (fabs(x - y) < eps)
10 public:
11     point solve(int np, point p[]) {//顶点数np,多边形顶点数组p,返回多边形的费马点
12         double nowx=0,nowy=0,nextx=0,nexty=0;
13         for(int i = 0; i < np; i++) {
14             nowx += p[i].x;
15             nowy += p[i].y;
16         }
17         for(nowx /= np, nowy /= np; ; nowx = nextx, nowy = nexty) {
18             double topx = 0, topy = 0, bot = 0;
19             for(int i = 0; i < np; i++) {
20                 double d = sqrt(Q(nowx -p[i].x) + Q(nowy - p[i].y));
21                 topx += p[i].x / d;
22                 topy += p[i].y / d;
23                 bot += 1 / d;
24             }
25             nextx = topx / bot;
26             nexty = topy / bot;
27             if(EQU(nextx, nowx) && EQU(nexty, nowy)) break;
28         }
29         point fp;
30         fp.x = nowx;
31         fp.y = nowy;
32         return fp;
33     }
34 } gx;
35 double Distance(point p1,point p2) {
36     return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
37 }
38 int main(){
39     while(~scanf("%lf%lf",&p[0].x,&p[0].y)){
40         for(int i=1;i<4;i++){
41             scanf("%lf%lf",&p[i].x,&p[i].y);
42         }
43         bool out=true;
44         for(int i=0;i<4;i++){
45             if(p[i].x!=-1||p[i].y!=-1){
46                 out=false;
47                 break;
48             }
49         }
50         if(out) break;
51         bool all=true;
52         for(int i=1;i<4;i++){
53             if(!EQU(p[i].x,p[0].x)||!EQU(p[i].y,p[0].y)){
54                 all=false;
55                 break;
56             }
57         }
58         if(all){
59             puts("0.0000");
60             continue;
61         }
62         point center=gx.solve(4,p);
63         double ans=0;
64         for(int i=0;i<4;i++){
65             ans+=Distance(center,p[i]);
66         }
67         printf("%.4f\n",ans);
68     }
69     return 0;
70 }

View Code

Computer Virus on Planet Pandora http://acm.hdu.edu.cn/showproblem.php?pid=3695

粘贴代码，有空重写一下。

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<algorithm>
  4 #include<cctype>
  5 #define mt(a,b) memset(a,b,sizeof(a))
  6 using namespace std;
  7 const int inf=0x3f3f3f3f;
  8 const int MAXPATLEN=1024;/** 模式串长度 */
  9 const int MAXPATNUM=256;/** 模式串数目 */
 10 const int MAXSTRLEN=5100010;/** 文本串长度 */
 11 class ACAuto {         /**多串匹配（AC自动机）*/
 12     static const int ALPHALEN = 26;
 13     static const int PATLEN = MAXPATLEN;
 14     static const int PATNUM = MAXPATNUM;
 15     static const int STRLEN = MAXSTRLEN;
 16     struct Node {
 17         int cnt;
 18         Node *fail,*chds[ALPHALEN];
 19         Node() : cnt(0), fail(NULL) {
 20             mt(chds,0);
 21         };
 22     }*root;
 23     int Hash(char a) {
 24         return a-'A';
 25     }
 26     void build_ACAuto() {
 27         Node **que = new Node*[PATNUM*PATLEN];
 28         int f = 0, r = 0;
 29         root->fail = root;
 30         for (int i = 0; i < ALPHALEN; i++) {
 31             Node *chd = root->chds[i];
 32             if (chd == NULL) {
 33                 root->chds[i] = root;
 34             } else {
 35                 chd->fail = root;
 36                 que[r++] = chd;
 37             }
 38         }
 39         while (f != r) {
 40             Node *now = que[f++];
 41             for (int i = 0; i < ALPHALEN; i++) {
 42                 Node *chd = now->chds[i];
 43                 if (chd != NULL) {
 44                     chd->fail = now->fail->chds[i];
 45                     que[r++] = chd;
 46                 } else {
 47                     now->chds[i] = now->fail->chds[i];
 48                 }
 49             }
 50         }
 51         delete [] que;
 52     }
 53     void Insert(char str[PATLEN]) {
 54         int str_len = strlen(str);
 55         Node *now = root;
 56         for (int i = 0; i < str_len; i++) {
 57             int cid = Hash(str[i]);
 58             if (now->chds[cid] == NULL) now->chds[cid] = new Node();
 59             now = now->chds[cid];
 60         }
 61         now->cnt++;
 62     }
 63 public:
 64     ACAuto(char pat_list[PATNUM][PATLEN], int list_len) {
 65         root = new Node();
 66         for (int i = 0; i < list_len; i++) {
 67             Insert(pat_list[i]);
 68         }
 69         build_ACAuto();
 70     }
 71     /** * 计算文本串中匹配所有模式串的次数之和（直接传入文本串） * @param 模式串 * @return 匹配次数 */
 72     int match_times(char str[STRLEN]) {
 73         int str_len = strlen(str);
 74         int times = 0;
 75         Node *now = root;
 76         for (int i = 0; i < str_len; i++) {
 77             now = now->chds[Hash(str[i])];
 78             times += now->cnt;
 79             Node *tmp = now;
 80             while (tmp != root) {
 81                 times += tmp->cnt;
 82                 tmp = tmp->fail;
 83             }
 84         }
 85         return times;
 86     }
 87     /** * 计算文本串中匹配所有模式串的次数之和（在函数内逐字符读入文本串） * @return 匹配次数 */
 88     int match_times() {
 89         int times = 0;
 90         char c;
 91         Node *now = root;
 92         while (c = getchar(), !isalpha(c));
 93         do {
 94             now = now->chds[Hash(c)];
 95             times += now->cnt;
 96             Node *tmp = now;
 97             while (tmp != root) {
 98                 times += tmp->cnt;
 99                 tmp = tmp->fail;
100             }
101         } while (c = getchar(), isalpha(c));
102         return times;
103     }
104     /** * 计算文本串中匹配到的模式串数目（直接传入文本串。将破坏结点的cnt信息！） * @return 匹配次数 */
105     int match_keynum(char str[STRLEN]) {
106         int str_len = strlen(str);
107         int times = 0;
108         Node *now = root;
109         for (int i = 0; i < str_len; i++) {
110             now = now->chds[Hash(str[i])];
111             Node *tmp = now;
112             while (tmp != root && tmp->cnt != -1) {
113                 times += tmp->cnt;
114                 tmp->cnt = -1;
115                 tmp = tmp->fail;
116             }
117         }
118         return times;
119     }
120     /** * 计算文本串中匹配到的模式串数目（在函数内逐字符读入文本串。将破坏结点 * 的cnt信息！） * @return 匹配次数 */
121     int match_keynum() {
122         int times = 0;
123         char c;
124         Node *now = root, *tmp;
125         while (c = getchar(), !isalpha(c));
126         do {
127             now = now->chds[Hash(c)];
128             tmp = now;
129             while (tmp != root && tmp->cnt != -1) {
130                 times += tmp->cnt;
131                 tmp->cnt = -1;
132                 tmp = tmp->fail;
133             }
134         } while (c = getchar(), isalpha(c));
135         return times;
136     }
137 };
138 char g[MAXPATNUM][MAXPATLEN];
139 char op[MAXSTRLEN];
140 char str[MAXSTRLEN];
141 int main() {
142     int t,n;
143     while(~scanf("%d",&t)) {
144         while(t--){
145             scanf("%d",&n);
146             for(int i=0; i<n; i++) {
147                 scanf("%s",g[i]);
148             }
149             scanf("%s",op);
150             int ls=0;
151             for(int i=0;op[i];){
152                 if(isalpha(op[i])){
153                     str[ls++]=op[i++];
154                     continue;
155                 }
156                 if(op[i]=='['){
157                     int sum=0;
158                     int j;
159                     for(j=i+1;op[j];j++){
160                         if(!isdigit(op[j])) break;
161                         sum*=10;
162                         sum+=op[j]-'0';
163                     }
164                     for(int k=0;k<sum;k++){
165                         str[ls++]=op[j];
166                     }
167                     i=j+2;
168                 }
169             }
170             int ans=0;
171             str[ls]=0;
172             ACAuto gx(g,n);
173             ans+=gx.match_keynum(str);
174             for(int i=0,j=strlen(str)-1;i<j;i++,j--){
175                 swap(str[i],str[j]);
176             }
177             ans+=gx.match_keynum(str);
178             printf("%d\n",ans);
179         }
180     }
181     return 0;
182 }

View Code

Selecting courses http://acm.hdu.edu.cn/showproblem.php?pid=3697

枚举时间加贪心选择终点靠前的，由于后台测试数据大于了题目描述，所以程序需具备鲁棒性。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define mt(a,b) memset(a,b,sizeof(a))
 5 using namespace std;
 6 const int M=512;
 7 struct point{
 8     int x,y;
 9     friend bool operator <(const point &a,const point &b){
10         return a.y<b.y;
11     }
12 }p[M];
13 bool use[M];
14 int main(){
15     int n;
16     while(~scanf("%d",&n),n){
17         int big=0;
18         for(int i=0;i<n;i++){
19             scanf("%d%d",&p[i].x,&p[i].y);
20             p[i].x*=2;
21             p[i].y*=2;
22             big=max(big,p[i].y);
23         }
24         sort(p,p+n);
25         int ans=0;
26         for(int s=0;s<10;s++){
27             mt(use,0);
28             for(int now=s;now<big;now+=10){
29                 for(int i=0;i<n;i++){
30                     if(!use[i]&&p[i].x<now&&now<p[i].y){
31                         use[i]=true;
32                         break;
33                     }
34                 }
35             }
36             int sum=0;
37             for(int i=0;i<n;i++){
38                 if(use[i]) sum++;
39             }
40             ans=max(ans,sum);
41         }
42         printf("%d\n",ans);
43     }
44     return 0;
45 }

View Code

end

posted on 2014-09-19 22:29 gaolzzxin 阅读(183) 评论(0) 编辑收藏举报