数据结构与算法实验题答案(2)

cin>>name_b;

if(ch=='e'||ch=='d') {

name_a.swap(name_b);//or swap(name_a,name_b) in ,but former is better. }

// cout<

if(relation==\ {

vector::iterator it=family_tree.begin(); for(;it!=family_tree.end();++it) {

if(it->name==name_b) break; }

int level_tmp=it->level;

if(it!=family_tree.end()) ++it; bool flag=1;

while(it!=family_tree.end()&&it->level>level_tmp) {

if(it->name==name_a&&it->level==level_tmp+2) {

cout<<\ flag=0; break; } ++it; }

if(flag) cout<<\ }

else if(relation==\ {

vector::iterator it=family_tree.begin(); for(;it!=family_tree.end();++it) {

if(it->name==name_a) break; }

int level_tmp=it->level;

if(it!=family_tree.end()) ++it; bool flag=1;

while(it!=family_tree.end()&&it->level>level_tmp) {

if(it->name==name_b) {

cout<<\ flag=0; break; } ++it; }

if(flag) cout<<\ } else {

vector::iterator it=family_tree.begin(); for(;it!=family_tree.end();++it) {

if(it->name==name_a) break; }

int level_tmp=it->level;

vector::iterator it_tmp=it; if(it!=family_tree.end()) ++it; bool flag=1;

while(it!=family_tree.end()&&it->level>=level_tmp) {

if(it->name==name_b&&it->level==level_tmp) {

cout<<\ flag=0; break; } ++it; }

if(flag) {

if(it_tmp!=family_tree.begin()) --it_tmp;

while(it_tmp!=family_tree.begin()&&it_tmp->level>=level_tmp) {

if(it_tmp->name==name_b&&it_tmp->level==level_tmp) {

cout<<\ flag=0; break; }

--it_tmp; }

}

if(flag) cout<<\ } } }

D 航空公司VIP客户查询（25）

源码：

#include #include

#include

#include

#include

#include

#include

#include

using namespace std;

typedef long int lld;//long int

#define rot(v) (((v)>>3)|(((v)&7)<<9))

const int MAX=110000;

const int MOD=100003;

const int INF=1000000000;

struct Node {

int a,b;

char x;

int mile,next;

}node[MAX];

int head[MOD],E;

char buf[22];

int chg(char x) {

if(x=='x')return 10;

return x-'0'; }

int fun(char s[],int &a,int &b,char &x) {

int ret=0,i;

int tmp;

a=b=0;

for(i=0;i<9;i++)

{

tmp=s[i]-'0';

a=a*10+tmp;//记录没取余数的数值

ret=ret*10+tmp;

}

ret%=MOD;

for(;i<17;i++)

{

tmp=s[i]-'0';

b=b*10+tmp;//记录后面没取余数的数值

ret=(ret*10+tmp)%MOD;

}

x=s[i];//记录最后一位

return (ret*100+chg(s[i]))%MOD;//返回取余数的数值 }

void ins(char s[],int mile) {

int a,b;

char x;

int h=fun(s,a,b,x);

int i;

//发生冲突时的处理

for(i=head[h];i!=-1;i=node[i].next)//head 全局变量

{

if(node[i].x==x&&node[i].a==a&&node[i].b==b)

{

node[i].mile+=mile;

return ;

}