HDU 3720 深搜 枚举

DES：从23个队员中选出4—4—2—1共4种11人来组成比赛队伍。给出每个人对每个职位的能力值。给出m组人在一起时会产生的附加效果。问你整场比赛人员的能力和最高是多少。

用深搜暴力枚举每种类型的人选择情况。感觉是这个深搜写的很机智。

在vector中存结构体就会很慢。TLE。直接存序号就AC了。以后还是尽量少用结构体吧。

#include<stdio.h>
#include<string.h>
#include<map>
#include<vector>
#include<iostream>
using namespace std;

vector<int>play[4];
vector<int>ans;
int pow[30];
map<string, int>mtype;
map<string, int>mname;
int eff[40][40];
int maxn = 0;

void dfs2(int, int );
void dfs3(int, int );
void dfs4(int, int );

void dfs1(int pos, int sum)
{
if (sum == 1)
{
dfs2(-1, 0);
return;
}
for (int i=pos+1; i<play[1].size(); ++i)
{
ans.push_back(play[1][i]);
dfs1(i, sum+1);
ans.pop_back();
}
}

void dfs2(int pos, int sum)
{
if (sum == 4)
{
dfs3(-1, 0);
return;
}
for (int i=pos+1; i<play[2].size(); ++i)
{
ans.push_back(play[2][i]);
dfs2(i, sum+1);
ans.pop_back();
}
}

void dfs3(int pos, int sum)
{
if (sum == 4)
{
dfs4(-1, 0);
return;
}
for (int i=pos+1; i<play[3].size(); ++i)
{
ans.push_back(play[3][i]);
dfs3(i, sum+1);
ans.pop_back();
}
}

void dfs4(int pos, int sum)
{
if (sum == 2)
{
int anss = 0;
for (int i=0; i<ans.size(); ++i)
{
for (int j=i+1; j<ans.size(); ++j)
{
anss += eff[ans[i]][ans[j]];
}
anss += pow[ans[i]];
}
if (anss > maxn)
maxn = anss;
return;
}
for (int i=pos+1; i<play[0].size(); ++i)
{
ans.push_back(play[0][i]);
dfs4(i, sum+1);
ans.pop_back();
}
}

int main()
{
mtype["goalkeeper"] = 1;
mtype["defender"] = 2;
mtype["midfielder"] = 3;
mtype["striker"] = 0;
char name[40], type[30];
int power;
while(cin >> name)
{
maxn = -0x3f3f3f3f;
ans.clear();
for (int i=0; i<4; ++i) play[i].clear();

cin >> power >> type;
int tt = mtype[type];
pow[0] = power;
mname[name] = 0;
play[tt].push_back(0);
memset(eff, 0, sizeof(eff));

for (int i=1; i<=22; ++i)
{
cin >> name >> power >> type;
pow[i] = power;
mname[name] = i;
int tt = mtype[type];
play[tt].push_back(i);
}

int n;
cin >> n;
int n1, n2, add;
char na1[40], na2[40];
for (int i=1; i<=n; ++i)
{
cin >> na1 >> na2 >> add;
n1 = mname[na1];
n2 = mname[na2];
eff[n1][n2] = add;
eff[n2][n1] = add;
}
int a[4] = {2, 1, 4, 4};
bool flag = true;
for (int i=0; i<4; ++i)
{
if (play[i].size() < a[i])
{
flag  = false;
break;
}
}
if (!flag)
{
cout << "impossible\n";
continue;
}

dfs1(-1, 0);
cout << maxn << endl;
}
return 0;
}

LOoK