UVA 1379 Pitcher Rotation

题目链接：http://acm.hust.edu.cn/vjudge/problem/41556

题意：有g+10场比赛，每天一场，每场都会对应一个对手。你有n个投手，每人一场比赛结束后都要休息四天，给出n个人对应m个对手的n*m胜率矩阵。求胜率的最大期望。

思路：每个人5天可以一轮换，所以说对于某个对手来讲，就派胜率前五的投手去对战即可。dp[i][a][b][c][d]表示第i天派d出赛，前三天分别是a，b，c的最大获胜期望。

0表示休息。枚举会麻烦一下而已，转移并不难。空间可以用滚动数组优化一下。

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <cstdlib>#include <iostream>#include <algorithm>#include <stack>#include <map>#include <set>#include <vector>#include <sstream>#include <queue>#include <utility>using namespace std;#define rep(i,j,k) for (int i=j;i<=k;i++)#define Rrep(i,j,k) for (int i=j;i>=k;i--)#define Clean(x,y) memset(x,y,sizeof(x))#define LL long long#define ULL unsigned long long#define inf 0x7fffffff#define mod 100000007int dp[2][6][6][6][6]; //滚动数组int k[250];int n,m,g;int a[109][109];vector<int> S[109];struct node{    int v;int pos;}b[109];bool cmp( node x , node y ){    return x.v > y.v;}void init(){    cin>>n>>m>>g;    S[0].clear();    S[0].push_back(0);    rep(i,1,m)    {        rep(j,1,n)        {            scanf("%d",&a[i][j]);            b[j].pos = j;            b[j].v = a[i][j];        }        sort(b+1,b+1+n,cmp);        S[i].clear(); // i 表示对手编号为i时，S[i]里面存的投手编号        rep(j,1,5) S[i].push_back( b[j].pos );    }    rep(i,1,g+10) cin>>k[i]; //比赛序列}void solve(){    Clean(dp,-1);    dp[0][0][0][0][0] = 0;    int last = 0 , now = 1;    int ta,tb,tc,td;    rep(i,1,g+10)    {        Clean(dp[now],-1);        ta = i>=2?k[i-1]:0;        tb = i>=3?k[i-2]:0;        tc = i>=4?k[i-3]:0;        td = i>=5?k[i-4]:0;        if ( k[i] != 0 ) //如果当天有比赛        {            for( int x = 0; x < (int)S[ k[i] ].size();x++ ) //枚举当天出赛的人            for( int t1 = 0; t1 < (int)S[ta].size(); t1++ ) //枚举前一天的            {                if ( S[ta][t1] == S[k[i]][x] ) continue;                for( int t2 = 0; t2 < (int)S[tb].size(); t2++ ) //枚举前两天的                {                    if ( S[tb][t2] == S[k[i]][x] ) continue;                    for( int t3 = 0; t3 < (int)S[tc].size(); t3++ ) //枚举前三天的                    {                        if ( S[tc][t3] == S[k[i]][x] ) continue;                        for( int t4 = 0; t4 < (int)S[td].size(); t4++ ) //枚举前四天的                        {                            if ( S[td][t4] == S[k[i]][x] ) continue;                            if ( dp[last][t4][t3][t2][t1] == -1 ) continue; //上一个状态求过 可以更新当前状态                            dp[now][t3][t2][t1][x] = max( dp[now][t3][t2][t1][x] , dp[last][t4][t3][t2][t1] + a[ k[i] ][ S[k[i]][x] ] );                        }                    }                }            }        }        else        {            for( int t1 = 0; t1 < 5; t1++ )                for( int t2 = 0; t2 < 5; t2++ )                    for( int t3 = 0; t3 < 5; t3++ )                        for( int t4 = 0; t4 < 5; t4++ )                            if ( dp[last][t4][t3][t2][t1] != -1 )                                dp[now][t3][t2][t1][0] = max( dp[now][t3][t2][t1][0] ,  dp[last][t4][t3][t2][t1] );        }        now ^= 1;        last ^= 1;    }    int ans = 0;    rep(i,0,4)        rep(j,0,4)            rep(k,0,4)                rep(l,0,4)                    ans = max( ans , dp[last][i][j][k][l] );    printf("%0.2f\n",ans/100.0);}int main(){    int T;    cin>>T;    while(T--)    {        init();        solve();    }    return 0;}