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[];
vector<int>ans;
int pow[];
map<string, int>mtype;
map<string, int>mname;
int eff[][];
int maxn = ;

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

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

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

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

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

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

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

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

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

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