HDU 3757 Evacuation Plan DP

跟 UVa 1474 - Evacuation Plan 一个题，但是在杭电上能交过，在UVa上交不过……不知道哪里有问题……

将施工队位置和避难所位置排序。

dp[i][j] 代表前 i 个避难所收留前 j 个施工队。

dp[i][j] = min( dp[i - 1][j - 1], dp[i][j - 1] ) + abs( b[i] - a[j] );

内存卡的比较死，要用滚动数组，并且记录路径的path[i][j]只能用bool型。MLE了四五次OTL……

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>

#define LL long long int

using namespace std;

const int MAXN = 4010;
const LL INF = (LL)1 << 50;

struct Team
{
    int id;
    LL pos;
};

LL dp[2][MAXN];
Team peo[MAXN];
Team shlt[MAXN];
int N, M;
bool path[MAXN][MAXN];
int ans[MAXN], getJ;

bool cmp( Team a, Team b )
{
    if ( a.pos != b.pos ) return a.pos < b.pos;
    return a.id < b.id;
}

LL DP()
{
    int pre = 0, cur = 1;

    for ( int i = 1; i <= N; ++i )
    {
        if ( N - i >= M - 1 )
            dp[1][i] = dp[1][i - 1] + abs( peo[i].pos - shlt[1].pos );
        else dp[1][i] = INF;
    }
    dp[1][0] = 0;

    for ( int i = 2; i <= M; ++i )
    {
        pre ^= 1;
        cur ^= 1;
        for ( int j = i; j <= N; ++j )
        {
            dp[cur][j] = dp[pre][j - 1] + abs( peo[j].pos - shlt[i].pos );
            path[i][j] = true;

            if ( j > i && dp[cur][j - 1] + abs( peo[j].pos - shlt[i].pos ) < dp[cur][j] )
            {
                dp[cur][j] = dp[cur][j - 1] + abs( peo[j].pos - shlt[i].pos );
                path[i][j] = false;
                
            }
            
        }
    }
    return dp[cur][N];
}

void PrintPath( int i, int j )
{
    if ( i == 1 )
    {
        getJ = j;
        return;
    }

    switch ( path[i][j] )
    {
    case false:
        PrintPath( i, j - 1 );
        break;
    case true:
        PrintPath( i - 1, j - 1 );
        break;
    }

    ans[ peo[j].id ] = shlt[i].id;

    return;
}

int main()
{
    while ( ~scanf( "%d", &N ) )
    {
        for ( int i = 1; i <= N; ++i )
        {
            scanf( "%lld", &peo[i].pos );
            peo[i].id = i;
        }
        sort( peo + 1, peo + N + 1, cmp );

        scanf( "%d", &M );
        for ( int j = 1; j <= M; ++j )
        {
            scanf( "%lld", &shlt[j].pos );
            shlt[j].id = j;
        }
        sort( shlt + 1, shlt + M + 1, cmp );

        printf( "%lld
", DP() );
        PrintPath( M, N );
        while ( getJ > 0 )
        {
            ans[ peo[ getJ ].id ] = shlt[1].id;
            --getJ;
        }

        for ( int i = 1; i <= N; ++i )
        {
            if ( i != 1 ) putchar(' ');
            printf( "%d", ans[i] );
        }
        puts("");
    }
    return 0;
}