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
";
            continue;
        }

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