TopCoder SRM 579 div2 练习
没参加比赛，就是在Arena中自己做了一下，还是感觉div 2和自己的水平比较匹配。。。。div 1也就能做第一题。
level 1，排序然后扫描即可
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
class PrimalUnlicensedCreatures {
public:
int maxWins(int, vector <int>);
};
int PrimalUnlicensedCreatures::maxWins(int initialLevel, vector <int> grezPower) {
int res = 0;
int len = grezPower.size();
sort(grezPower.begin(), grezPower.end());
int power = initialLevel;
for(int i=0; i<len; i++){
if(power > grezPower[i]){
res++;
power += grezPower[i]/2;
}
else
break;
}
return res;
}
//<%:testing-code%>
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level 2:陷阱很多啊，要仔细看样例，字符串的处理中，需要看undo history还有前一个字符串是否是当前字符串的子串，要注意：没有删除功能，必须双击。
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
class UndoHistory {
public:
int minPresses(vector <string>);
int presub(string a, string b){
int r = min(a.size(), b.size());
int ans = 0;
for(int i=0; i<r; i++){
if(a[i] != b[i])
return i;
}
return r;
}
bool check(string master, string slave){
if(master.size() < slave.size())
return false;
for(int i=0; i<slave.size(); i++)
if(slave[i] != master[i])
return false;
return true;
}
};
int UndoHistory::minPresses(vector <string> lines) {
int len = lines.size();
int res = len;
int pre[55];
memset(pre, 0, sizeof(pre));
for(int i=0; i<len; i++)
for(int j=i-1; j>=0; j--)
pre[i] = max(pre[i], presub(lines[j], lines[i]));
for(int i=0; i<len; i++){
if(i!=0 && check(lines[i], lines[i-1])==true)
res += min(lines[i].size()-lines[i-1].size(), 2+lines[i].size()-pre[i]);
else if (i!=0)
res += 2+lines[i].size()-pre[i];
else
res += lines[i].size();
}
return res;
}
//<%:testing-code%>
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level 3：几何题目，不过也就只能算是初中的几何，两个圆相切，求出圆心横坐标的距离。一个tricky的地方：第i个圆不一定和第i-1个圆相切，考虑两个圆的半径相差特别大的时候就可以明白这一点，嗯。
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
class MarblePositioning {
public:
int len;
double res;
bool used[10];
vector<int> r;
double record[10];
double totalWidth(vector <int>);
void solve(double cur, int n, int pre){
if(n==len){
res = min(res, cur);
return;
}
for(int i=0; i<len; i++){
if(used[i] == true)
continue;
used[i]=true;
double tmp = 0;
for(int j=0; j<len; j++){
if(used[j]==false || i==j)
continue;
tmp = max(tmp, record[j]+2.0*sqrt(1.0*r[i]*r[j]));
}
record[i] = tmp;
solve(tmp, n+1, i);
used[i]=false;
}
return;
}
};
double MarblePositioning::totalWidth(vector <int> radius) {
len = radius.size();
r = radius;
res = 999999999999LL;
memset(used, false, sizeof(used));
memset(record, 0, sizeof(record));
solve(0, 0, -1);
return res;
}
//<%:testing-code%>
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