商人过河问题（DFS）

来源：互联网发布：Java仿hashmap实现编辑：程序博客网时间：2024/05/18 03:21

问题描述：
3个商人带着3个仆人过河，过河的工具只有一艘小船，只能同时载两个人过河，包括划船的人。在河的任何一边，只要仆人的数量超过商人的数量，仆人就会联合起来将商人杀死并抢夺其财物，问商人应如何设计过河顺序才能让所有人都安全地过到河的另一边。

详细过程参见《数学模型》第四版（姜启源）

#include <cstdio>#define maxn 101int num;//number of bus or folint graph[maxn*maxn][maxn*maxn];int state[maxn][maxn];//when cross riverint c_bus[5] = {2, 1, 0, 1, 0};int c_fol[5] = {0, 1, 2, 0, 1};int b_step[maxn*maxn];int f_step[maxn*maxn];bool flag = false;void DFS(int bus, int fol, int step, int dir){    b_step[step] = bus, f_step[step] = fol;    if(bus == 0 && fol == 0)    {        for(int i = 0; i <= step; i++)        {            printf("(%d,%d)", b_step[i], f_step[i]);            if(i != step )                printf(" -> ");        }        printf("\n");        flag = true;    }    int fa = bus * ( num + 1 ) + fol;    for(int i = 0; i < 5; i++)    {        if(dir){            int b_next = bus - c_bus[i], f_next = fol - c_fol[i];            if(b_next >= 0 && b_next < num+1 && f_next >= 0 && f_next < num + 1 && state[b_next][f_next])            {                int son = b_next * ( num + 1 ) + f_next;                if(!graph[fa][son] && !graph[son][fa])                {                    graph[fa][son] = 1;                    graph[son][fa] = 1;                    DFS(b_next, f_next, step + 1, !dir);                    graph[fa][son] = 0;                    graph[fa][son] = 0;                }            }        }        else{            int b_next = bus + c_bus[i], f_next = fol + c_fol[i];            if(b_next >= 0 && b_next < num + 1 && f_next >= 0 && f_next < num + 1 && state[b_next][f_next])            {                int son = b_next * ( num + 1) + f_next;                if(!graph[fa][son] && !graph[son][fa])                {                    graph[fa][son] = 1;                    graph[son][fa] = 1;                    DFS(b_next, f_next, step + 1, !dir);                    graph[fa][son] = 0;                    graph[fa][son] = 0;                }            }        }    }}int main(){    printf("Please input the number of the businessman: ");    scanf("%d",&num);    for(int i = 0; i < num + 1; i++)    {        state[i][0] = 1;        state[i][num] = 1;        state[i][i] = 1;    }    DFS(num, num, 0, 1);    if(!flag)        printf("they can't cross the river.");}

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