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……

 

  1 #include <cstdio>

  2 #include <cstring>

  3 #include <cstdlib>

  4 #include <algorithm>

  5 

  6 #define LL long long int

  7 

  8 using namespace std;

  9 

 10 const int MAXN = 4010;

 11 const LL INF = (LL)1 << 50;

 12 

 13 struct Team

 14 {

 15     int id;

 16     LL pos;

 17 };

 18 

 19 LL dp[2][MAXN];

 20 Team peo[MAXN];

 21 Team shlt[MAXN];

 22 int N, M;

 23 bool path[MAXN][MAXN];

 24 int ans[MAXN], getJ;

 25 

 26 bool cmp( Team a, Team b )

 27 {

 28     if ( a.pos != b.pos ) return a.pos < b.pos;

 29     return a.id < b.id;

 30 }

 31 

 32 LL DP()

 33 {

 34     int pre = 0, cur = 1;

 35 

 36     for ( int i = 1; i <= N; ++i )

 37     {

 38         if ( N - i >= M - 1 )

 39             dp[1][i] = dp[1][i - 1] + abs( peo[i].pos - shlt[1].pos );

 40         else dp[1][i] = INF;

 41     }

 42     dp[1][0] = 0;

 43 

 44     for ( int i = 2; i <= M; ++i )

 45     {

 46         pre ^= 1;

 47         cur ^= 1;

 48         for ( int j = i; j <= N; ++j )

 49         {

 50             dp[cur][j] = dp[pre][j - 1] + abs( peo[j].pos - shlt[i].pos );

 51             path[i][j] = true;

 52 

 53             if ( j > i && dp[cur][j - 1] + abs( peo[j].pos - shlt[i].pos ) < dp[cur][j] )

 54             {

 55                 dp[cur][j] = dp[cur][j - 1] + abs( peo[j].pos - shlt[i].pos );

 56                 path[i][j] = false;

 57                 

 58             }

 59             

 60         }

 61     }

 62     return dp[cur][N];

 63 }

 64 

 65 void PrintPath( int i, int j )

 66 {

 67     if ( i == 1 )

 68     {

 69         getJ = j;

 70         return;

 71     }

 72 

 73     switch ( path[i][j] )

 74     {

 75     case false:

 76         PrintPath( i, j - 1 );

 77         break;

 78     case true:

 79         PrintPath( i - 1, j - 1 );

 80         break;

 81     }

 82 

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

 84 

 85     return;

 86 }

 87 

 88 int main()

 89 {

 90     while ( ~scanf( "%d", &N ) )

 91     {

 92         for ( int i = 1; i <= N; ++i )

 93         {

 94             scanf( "%lld", &peo[i].pos );

 95             peo[i].id = i;

 96         }

 97         sort( peo + 1, peo + N + 1, cmp );

 98 

 99         scanf( "%d", &M );

100         for ( int j = 1; j <= M; ++j )

101         {

102             scanf( "%lld", &shlt[j].pos );

103             shlt[j].id = j;

104         }

105         sort( shlt + 1, shlt + M + 1, cmp );

106 

107         printf( "%lld\n", DP() );

108         PrintPath( M, N );

109         while ( getJ > 0 )

110         {

111             ans[ peo[ getJ ].id ] = shlt[1].id;

112             --getJ;

113         }

114 

115         for ( int i = 1; i <= N; ++i )

116         {

117             if ( i != 1 ) putchar(' ');

118             printf( "%d", ans[i] );

119         }

120         puts("");

121     }

122     return 0;

123 }