UVA 116 Unidirectional TSP

UVA_116

    这个题目可以用记忆化搜索实现，由于矩阵中的数字不全为正，故可以把INF作为还没有搜索过的标记。由于要求字典序最小，所以分情况控制好搜索顺序即可。

#include<stdio.h>
#include<string.h>
#define INF 1000000000
int R, C, a[20][110], f[20][110], next[2200];
void init()
{
    int i, j;
    for(i = 0; i < R; i ++)
        for(j = 0; j < C; j ++)
            scanf("%d", &a[i][j]);
}
int dfs(int x, int y)
{
    int i, j, k, newx, min;
    if(f[x][y] != INF)
        return f[x][y];
    min = INF;
    if(x == 0)
    {
        for(i = 0; i < 2; i ++)
        {
            if(x + i >= R)
                break;
            newx = x + i;
            k = dfs(newx, y + 1);
            if(k < min)
            {
                min = k;
                next[x * C + y] = newx * C + y + 1;
            }
        }
        newx = R - 1;
        k = dfs(newx, y + 1);
        if(k < min)
        {
            min = k;
            next[x * C + y] = newx * C + y + 1;
        }
    }
    else if(x == R - 1)
    {
        newx = 0;
        k = dfs(newx, y + 1);
        if(k < min)
        {
            min = k;
            next[x * C + y] = newx * C + y + 1;
        }
        for(i = -1; i < 1; i ++)
        {
            newx = x + i;
            k = dfs(newx, y + 1);
            if(k < min)
            {
                min = k;
                next[x * C + y] = newx * C + y + 1;
            }
        }
    }
    else
    {
        for(i = -1; i < 2; i ++)
        {
            newx = x + i;
            k = dfs(newx, y + 1);
            if(k < min)
            {
                min = k;
                next[x * C + y] = newx * C + y + 1;
            }
        }
    }
    return f[x][y] = min + a[x][y];
}
void printresult(int z, int flag)
{
    if(flag)
        printf("");
    else
        flag = 1;
    printf("%d", z / C + 1);
    if(next[z] != -1)
        printresult(next[z], flag);
}
void solve()
{
    int i, j, k, min, z;
    for(i = 0; i < R; i ++)
        for(j = 0; j < C; j ++)
            f[i][j] = INF;
    memset(next, -1, sizeof(next));
    for(i = 0; i < R; i ++)
        f[i][C - 1] = a[i][C - 1];
    min = INF;
    for(i = 0; i < R; i ++)
    {
        k = dfs(i, 0);
        if(k < min)
        {
            min = k;
            z = i * C;
        }
    }
    printresult(z, 0);
    printf("\n");
    printf("%d\n", min);
}
int main()
{
    while(scanf("%d%d", &R, &C) == 2)
    {
        init();
        solve();
    }
    return 0;
}