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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 101
int num;//number of bus or fol
int graph[maxn*maxn][maxn*maxn];
int state[maxn][maxn];
//when cross river
int 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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原文地址：http://www.cnblogs.com/yqbeyond/p/4391287.html
