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  • POJ 2374 Fence Obstacle Course

    题目链接:http://poj.org/problem?id=2374

    DP+线段树

    View Code
      1 #include <cstdio>
      2 #include <cstring>
      3 #include <cstdlib>
      4 #include <algorithm>
      5 
      6 #define lson l, m, rt << 1
      7 #define rson m + 1, r, rt << 1 | 1
      8 
      9 using namespace std;
     10 
     11 const int MAXN = 100000;
     12 const int MAXSIZE = MAXN << 1;
     13 const int INF = 2147483645;
     14 
     15 struct Fence
     16 {
     17     int st, ed;
     18 };
     19 
     20 int N, S;
     21 Fence F[ MAXN + 10 ];
     22 int up[ ( MAXSIZE << 2 ) + 10 ], flag[ ( MAXSIZE << 2 ) + 10 ];
     23 int dp[50010][2];
     24 int MinBound, MaxBound;
     25 
     26 void PushDown( int rt )
     27 {
     28     int lc = rt << 1;
     29     int rc = rt << 1 | 1;
     30     if ( flag[rt] != -1 )
     31     {
     32         flag[lc] = flag[rc] = flag[rt];
     33         up[lc] = up[rc] = flag[rt];
     34         flag[rt] = -1;
     35     }
     36     return;
     37 }
     38 
     39 void Update( int L, int R, int c, int l, int r, int rt )
     40 {
     41     if ( L <= l && r <= R )
     42     {
     43         up[rt] = c;
     44         flag[rt] = c;
     45         return;
     46     }
     47     PushDown( rt );
     48 
     49     int m = ( l + r ) >> 1;
     50     if ( L <= m ) Update( L, R, c, lson );
     51     if ( R > m ) Update( L, R, c, rson );
     52 
     53     return;
     54 }
     55 
     56 int Query( int L, int l, int r, int rt )
     57 {
     58     if ( L == l && L == r )
     59     {
     60         return up[rt];
     61     }
     62     PushDown(rt);
     63 
     64     int m = ( l + r ) >> 1;
     65     if ( L <= m ) return Query( L, lson );
     66     else return Query( L, rson );
     67 }
     68 
     69 void Solved()
     70 {
     71     memset( dp[0], 0, sizeof(dp[0]) );
     72 
     73     for ( int i = 1; i <= N; ++i )
     74     {
     75         int left = Query( F[i].st, MinBound, MaxBound, 1 );
     76         int right = Query( F[i].ed, MinBound, MaxBound, 1 );
     77         //printf( "left=%d right=%d\n", left, right );
     78 
     79         dp[i][0] = min( dp[left][0] + abs( F[left].st - F[i].st ) , dp[left][1] + abs( F[left].ed - F[i].st ) );
     80         dp[i][1] = min( dp[right][0] + abs( F[right].st - F[i].ed ) , dp[right][1] + abs( F[right].ed - F[i].ed ) );
     81 
     82         Update( F[i].st, F[i].ed, i, MinBound, MaxBound, 1 );
     83     }
     84 
     85     int ans = min( dp[N][0] + abs( F[N].st - S - MAXN ), dp[N][1] + abs( F[N].ed - S - MAXN ) );
     86     printf( "%d\n", ans );
     87 
     88     return;
     89 }
     90 
     91 int main()
     92 {
     93     while ( ~scanf( "%d%d", &N, &S ) )
     94     {
     95         MinBound = INF;
     96         MaxBound = -1;
     97 
     98         F[0].st = 0 + MAXN;
     99         F[0].ed = 0 + MAXN;
    100 
    101         for ( int i = 1; i <= N; ++i )
    102         {
    103             scanf( "%d%d", &F[i].st, &F[i].ed );
    104             F[i].st += MAXN;
    105             F[i].ed += MAXN;
    106             MinBound = min( MinBound, F[i].st );
    107             MaxBound = max( MaxBound, F[i].ed );
    108         }
    109 
    110         memset( up, 0, sizeof(up) );
    111         memset( flag, -1, sizeof(flag) );
    112 
    113         Solved();
    114     }
    115     return 0;
    116 }
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  • 原文地址:https://www.cnblogs.com/GBRgbr/p/3040306.html
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