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。

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<map>
 #define mt(a,b) memset(a,b,sizeof(a))
 using namespace std;
 const int M=;
 char sa[M],sb[M],sc[M],ha[M],hb[M],op[]="+-*/";
 int val[];
 bool use[];
 map<string,bool> mp;
 string str;
 bool prezero(char s[]){
     if(strlen(s)>&&!val[s[]-'A']) return true; return false;
 }
 int getint(char s[]){
     int res=;
     for(int i=;s[i];i++){
         res=(res<<)+(res<<)+val[s[i]-'A'];
     }
     return res;
 }
 int solve(const int &a,const char &ope,const int &b){
     if(ope=='+') return a+b;
     if(ope=='-') return a-b;
     if(ope=='*') return a*b;
     if(b&&!(a%b))return a/b;
     return -;
 }
 void flag(const int &a,const char &ope,const int &b){
     sprintf(ha,"%d",a);
     sprintf(hb,"%d",b);
     str=(string)ha+ope+(string)hb;
     mp[str]=true;
 }
 void dfs(int t){
     if(t==){
         if(prezero(sa)||prezero(sb)||prezero(sc)) return ;
         int a=getint(sa);
         int b=getint(sb);
         int c=getint(sc);
         for(int i=;i<;i++){
             if(solve(a,op[i],b)==c){
                 flag(a,op[i],b);
             }
         }
         return ;
     }
     for(int i=;i<;i++){
         if(!use[i]){
             val[t]=i;
             use[i]=true;
             dfs(t+);
             use[i]=false;
         }
     }
 }
 int main(){
     int t;
     while(~scanf("%d",&t)){
         while(t--){
             scanf("%s%s%s",sa,sb,sc);
             mt(use,);
             mp.clear();
             dfs();
             printf("%d\n",mp.size());
         }
     }
     return ;
 }

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。

 #include<cstdio>
 #include<cmath>
 const double eps=1e-;
 struct point{
     double x,y;
 }p[];
 class Fermatpoint {//多边形的费马点
 #define Q(x) ((x) * (x))
 #define EQU(x, y) (fabs(x - y) < eps)
 public:
     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
         double nowx=,nowy=,nextx=,nexty=;
         for(int i = ; i < np; i++) {
             nowx += p[i].x;
             nowy += p[i].y;
         }
         for(nowx /= np, nowy /= np; ; nowx = nextx, nowy = nexty) {
             double topx = , topy = , bot = ;
             for(int i = ; i < np; i++) {
                 double d = sqrt(Q(nowx -p[i].x) + Q(nowy - p[i].y));
                 topx += p[i].x / d;
                 topy += p[i].y / d;
                 bot +=  / d;
             }
             nextx = topx / bot;
             nexty = topy / bot;
             if(EQU(nextx, nowx) && EQU(nexty, nowy)) break;
         }
         point fp;
         fp.x = nowx;
         fp.y = nowy;
         return fp;
     }
 } gx;
 double Distance(point p1,point p2) {
     return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
 }
 int main(){
     while(~scanf("%lf%lf",&p[].x,&p[].y)){
         for(int i=;i<;i++){
             scanf("%lf%lf",&p[i].x,&p[i].y);
         }
         bool out=true;
         for(int i=;i<;i++){
             if(p[i].x!=-||p[i].y!=-){
                 out=false;
                 break;
             }
         }
         if(out) break;
         bool all=true;
         for(int i=;i<;i++){
             if(!EQU(p[i].x,p[].x)||!EQU(p[i].y,p[].y)){
                 all=false;
                 break;
             }
         }
         if(all){
             puts("0.0000");
             continue;
         }
         point center=gx.solve(,p);
         double ans=;
         for(int i=;i<;i++){
             ans+=Distance(center,p[i]);
         }
         printf("%.4f\n",ans);
     }
     return ;
 }

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

粘贴代码，有空重写一下。

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cctype>
 #define mt(a,b) memset(a,b,sizeof(a))
 using namespace std;
 const int inf=0x3f3f3f3f;
 const int MAXPATLEN=;/** 模式串长度 */
 const int MAXPATNUM=;/** 模式串数目 */
 const int MAXSTRLEN=;/** 文本串长度 */
 class ACAuto {         /**多串匹配（AC自动机）*/
     static const int ALPHALEN = ;
     static const int PATLEN = MAXPATLEN;
     static const int PATNUM = MAXPATNUM;
     static const int STRLEN = MAXSTRLEN;
     struct Node {
         int cnt;
         Node *fail,*chds[ALPHALEN];
         Node() : cnt(), fail(NULL) {
             mt(chds,);
         };
     }*root;
     int Hash(char a) {
         return a-'A';
     }
     void build_ACAuto() {
         Node **que = new Node*[PATNUM*PATLEN];
         int f = , r = ;
         root->fail = root;
         for (int i = ; i < ALPHALEN; i++) {
             Node *chd = root->chds[i];
             if (chd == NULL) {
                 root->chds[i] = root;
             } else {
                 chd->fail = root;
                 que[r++] = chd;
             }
         }
         while (f != r) {
             Node *now = que[f++];
             for (int i = ; i < ALPHALEN; i++) {
                 Node *chd = now->chds[i];
                 if (chd != NULL) {
                     chd->fail = now->fail->chds[i];
                     que[r++] = chd;
                 } else {
                     now->chds[i] = now->fail->chds[i];
                 }
             }
         }
         delete [] que;
     }
     void Insert(char str[PATLEN]) {
         int str_len = strlen(str);
         Node *now = root;
         for (int i = ; i < str_len; i++) {
             int cid = Hash(str[i]);
             if (now->chds[cid] == NULL) now->chds[cid] = new Node();
             now = now->chds[cid];
         }
         now->cnt++;
     }
 public:
     ACAuto(char pat_list[PATNUM][PATLEN], int list_len) {
         root = new Node();
         for (int i = ; i < list_len; i++) {
             Insert(pat_list[i]);
         }
         build_ACAuto();
     }
     /** * 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 */
     int match_times(char str[STRLEN]) {
         int str_len = strlen(str);
         int times = ;
         Node *now = root;
         for (int i = ; i < str_len; i++) {
             now = now->chds[Hash(str[i])];
             times += now->cnt;
             Node *tmp = now;
             while (tmp != root) {
                 times += tmp->cnt;
                 tmp = tmp->fail;
             }
         }
         return times;
     }
     /** * 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 */
     int match_times() {
         int times = ;
         char c;
         Node *now = root;
         while (c = getchar(), !isalpha(c));
         do {
             now = now->chds[Hash(c)];
             times += now->cnt;
             Node *tmp = now;
             while (tmp != root) {
                 times += tmp->cnt;
                 tmp = tmp->fail;
             }
         } while (c = getchar(), isalpha(c));
         return times;
     }
     /** * 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 */
     int match_keynum(char str[STRLEN]) {
         int str_len = strlen(str);
         int times = ;
         Node *now = root;
         for (int i = ; i < str_len; i++) {
             now = now->chds[Hash(str[i])];
             Node *tmp = now;
             while (tmp != root && tmp->cnt != -) {
                 times += tmp->cnt;
                 tmp->cnt = -;
                 tmp = tmp->fail;
             }
         }
         return times;
     }
     /** * 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 */
     int match_keynum() {
         int times = ;
         char c;
         Node *now = root, *tmp;
         while (c = getchar(), !isalpha(c));
         do {
             now = now->chds[Hash(c)];
             tmp = now;
             while (tmp != root && tmp->cnt != -) {
                 times += tmp->cnt;
                 tmp->cnt = -;
                 tmp = tmp->fail;
             }
         } while (c = getchar(), isalpha(c));
         return times;
     }
 };
 char g[MAXPATNUM][MAXPATLEN];
 char op[MAXSTRLEN];
 char str[MAXSTRLEN];
 int main() {
     int t,n;
     while(~scanf("%d",&t)) {
         while(t--){
             scanf("%d",&n);
             for(int i=; i<n; i++) {
                 scanf("%s",g[i]);
             }
             scanf("%s",op);
             int ls=;
             for(int i=;op[i];){
                 if(isalpha(op[i])){
                     str[ls++]=op[i++];
                     continue;
                 }
                 if(op[i]=='['){
                     int sum=;
                     int j;
                     for(j=i+;op[j];j++){
                         if(!isdigit(op[j])) break;
                         sum*=;
                         sum+=op[j]-'';
                     }
                     for(int k=;k<sum;k++){
                         str[ls++]=op[j];
                     }
                     i=j+;
                 }
             }
             int ans=;
             str[ls]=;
             ACAuto gx(g,n);
             ans+=gx.match_keynum(str);
             for(int i=,j=strlen(str)-;i<j;i++,j--){
                 swap(str[i],str[j]);
             }
             ans+=gx.match_keynum(str);
             printf("%d\n",ans);
         }
     }
     return ;
 }

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。

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define mt(a,b) memset(a,b,sizeof(a))
 using namespace std;
 const int M=;
 struct point{
     int x,y;
     friend bool operator <(const point &a,const point &b){
         return a.y<b.y;
     }
 }p[M];
 bool use[M];
 int main(){
     int n;
     while(~scanf("%d",&n),n){
         int big=;
         for(int i=;i<n;i++){
             scanf("%d%d",&p[i].x,&p[i].y);
             p[i].x*=;
             p[i].y*=;
             big=max(big,p[i].y);
         }
         sort(p,p+n);
         int ans=;
         for(int s=;s<;s++){
             mt(use,);
             for(int now=s;now<big;now+=){
                 for(int i=;i<n;i++){
                     if(!use[i]&&p[i].x<now&&now<p[i].y){
                         use[i]=true;
                         break;
                     }
                 }
             }
             int sum=;
             for(int i=;i<n;i++){
                 if(use[i]) sum++;
             }
             ans=max(ans,sum);
         }
         printf("%d\n",ans);
     }
     return ;
 }

end