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            Lake Counting
            Time Limit:1000MS? Memory Limit:65536K
            Total Submit:1360 Accepted:629

            Description
            Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.

            Given a diagram of Farmer John's field, determine how many ponds he has.

            Input
            * Line 1: Two space-separated integers: N and M

            * Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.

            Output
            * Line 1: The number of ponds in Farmer John's field.

            Sample Input

            10 12
            W........WW.
            .WWW.....WWW
            ....WW...WW.
            .........WW.
            .........W..
            ..W......W..
            .W.W.....WW.
            W.W.W.....W.
            .W.W......W.
            ..W.......W.

            Sample Output

            3

            Hint
            OUTPUT DETAILS:

            There are three ponds: one in the upper left, one in the lower left,and one along the right side.

            Source
            USACO 2004 November

            #include? < iostream >
            using ? namespace ?std;

            const ? int ?MAX? = ? 102 ;
            int ?g[MAX][MAX];

            void ?travel( int ?i,? int ?j)
            {
            ????
            int ?incr[ 8 ][ 2 ]? = ? { { 0 , - 1 } ,? { 1 , - 1 } ,? { 1 , 0 } ,? { 1 , 1 } ,? { 0 , 1 } ,? { - 1 , 1 } ,? { - 1 , 0 } ,? { - 1 , - 1 } } ;
            ????
            int ?k;
            ????
            int ?tmpI,?tmpJ;
            ????
            for ?(k = 0 ;?k < 8 ;?k ++ )? {
            ????????tmpI?
            = ?i? + ?incr[k][ 0 ];
            ????????tmpJ?
            = ?j? + ?incr[k][ 1 ];
            ????????
            if ?(g[tmpI][tmpJ]? == ? 1 )? {
            ????????????g[tmpI][tmpJ]?
            = ? - 1 ;
            ????????????travel(tmpI,?tmpJ);
            ????????}

            ????}

            }


            int ?main()
            {
            ????
            int ?n,?m;
            ????
            int ?i,?j;
            ????
            int ?ans? = ? 0 ;
            ????
            char ?t;
            ????cin?
            >> ?n? >> ?m;
            ????
            for ?(i = 0 ;?i <= m + 1 ;?i ++ )? {
            ????????g[
            0 ][i]? = ? - 1 ;
            ????????g[n
            + 1 ][i]? = ? - 1 ;
            ????}

            ????
            for ?(i = 0 ;?i <= n + 1 ;?i ++ )? {
            ????????g[i][
            0 ]? = ? - 1 ;
            ????????g[i][m
            + 1 ]? = ? - 1 ;
            ????}

            ????
            for ?(i = 1 ;?i <= n;?i ++ )? {
            ????????
            for ?(j = 1 ;?j <= m;?j ++ )? {
            ????????????cin?
            >> ?t;
            ????????????
            if ?(t? == ? ' W ' )? {
            ????????????????g[i][j]?
            = ? 1 ;
            ????????????}
            ? else ? {
            ????????????????g[i][j]?
            = ? 0 ;
            ????????????}

            ????????}

            ????}

            ????
            for ?(i = 1 ;?i <= n;?i ++ )? {
            ????????
            for ?(j = 1 ;?j <= m;?j ++ )? {
            ????????????
            if ?(g[i][j]? == ? 1 )? {
            ????????????????ans
            ++ ;
            ????????????????g[i][j]?
            = ? - 1 ;
            ????????????????travel(i,?j);
            ????????????}

            ????????}

            ????}

            ????cout?
            << ?ans? << ?endl;
            ????
            return ? 0 ;
            }
            感受遞歸的神奇!
            posted on 2006-05-01 18:48 閱讀(712) 評(píng)論(0)  編輯 收藏 引用 所屬分類: ACM題目
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