## 2013年10月12日 星期六

### UVa 116 - Unidirectional TSP

Keyword: DP, greedy

Algorithm:

A brute force will require 3^100 methods, which is too large to handle, so we must retreat to DP method.
First let's say that an "answer array" stores the lowest cost of going from the cell to the end.
Started from the last column, the path with lowest cost starting from each row is the cell it self because only one step is needed to get to the end.
Then deal with next column, each row will have 3 choices.
To get the lowest cost, in each row, we have to select the choice with lowest cost.
If 2 choices result in the same cost, choose the one with lower lexicographically order.
After this operation, this column will show the minimum cost starting from each row to the end.
Then deals with next column until the begin column.
Finally, search the lowest cost and iterate through the path.

Code:
`#include<cstdio>#include<cstring>#include<algorithm>template<typename T>inline T min( T a, T b, T c ) { return std::min( std::min( a, b ), c ); }int main(){    int h, w, num[ 10 ][ 100 ], next[ 10 ][ 100 ], sum[ 10 ][ 100 ];    while( scanf( "%d %d", &h, &w ) != EOF )    {        int tar = 0, MIN = 1000000000;        memset( next, 0, sizeof( next ) );        memset( sum, 0, sizeof( sum ) );        for( int i = 0; i < h; ++i )            for( int j = 0; j < w; ++j )                scanf( "%d", num[ i ]+j );        for( int i = 0; i < h; ++i )            sum[ i ][ w-1 ] = num[ i ][ w-1 ];        for( int i = w-2; i >= 0; --i )            for( int j = 0; j < h; ++j )            if( j == 0 )            {                sum[ j ][ i ] = min( sum[ h-1 ][ i+1 ], sum[ 0 ][ i+1 ], h > 1? sum[ 1 ][ i+1 ] : sum[ 0 ][ i+1 ] );                if( sum[ j ][ i ] == sum[ 0 ][ i+1 ] )                    next[ j ][ i ] = 0;                else if( sum[ j ][ i ] == sum[ 1 ][ i+1 ] )                    next[ j ][ i ] = 1;                else                    next[ j ][ i ] = h-1;                sum[ j ][ i ] += num[ j ][ i ];            }            else if( j == h-1 )            {                sum[ j ][ i ] = min( sum[ 0 ][ i+1 ], sum[ j ][ i+1 ], sum[ j-1 ][ i+1 ] );                if( sum[ j ][ i ] == sum[ 0 ][ i+1 ] )                    next[ j ][ i ] = 0;                else if( sum[ j ][ i ] == sum[ j-1 ][ i+1 ] )                    next[ j ][ i ] = j-1;                else                    next[ j ][ i ] = j;                sum[ j ][ i ] += num[ j ][ i ];            }            else            {                sum[ j ][ i ] = min( sum[ j-1 ][ i+1 ], sum[ j ][ i+1 ], sum[ j+1 ][ i+1 ] );                if( sum[ j ][ i ] == sum[ j-1 ][ i+1 ] )                    next[ j ][ i ] = j-1;                else if( sum[ j ][ i ] == sum[ j ][ i+1 ] )                    next[ j ][ i ] = j;                else                    next[ j ][ i ] = j+1;                sum[ j ][ i ] += num[ j ][ i ];            }        for( int i = 0; i < h; ++i )            if( MIN > sum[ i ][ 0 ] )                MIN = sum[ i ][ 0 ], tar = i;        for( int i = 0; i < w; ++i )            printf( "%d%c", tar+1, i == w-1? '\n' : ' ' ), tar = next[ tar ][ i ];        printf( "%d\n", MIN );    }    return 0;}`