poj 3311 floyd + 状态

题目链接：http://poj.org/problem?id=3311

先Floyd处理了，变成tsp问题

if(k == 0)

tem = 0;

else tem = (1<<(k-1));//因为点是从1开始的，懒得预处理dp[0][i]

dp[i+tem][k] = min(dp[i+tem][k],dp[i][j]+map[j][k]);//由这次状态写出下次的

#include <stdio.h>
#include <string.h>
#define inf 0x7f7f7f7f

int n;
int map[11][11];
int dp[1<<12][12];

int min(int a,int b)
{
if(a > b)
return b;
return a;
}

int main()
{
int i,j,k;
int n,tem,tmp;

while(~scanf("%d",&n),n)//因为没加n=0，wa七次，哭晕在厕所
{

memset(dp,inf,sizeof(dp));

for(i = 0; i <= n; i++)
{
for(j = 0; j <= n; j++)
{
scanf("%d",&map[i][j]);
}
}

for(k = 0; k <= n; k++)
{
for(i = 0; i <= n; i++)
{
for(j = 0; j <= n; j++)
{
map[i][j] = min(map[i][j],map[i][k]+map[k][j]);
}
}
}

dp[0][0] = 0;
tmp = 1<<n;
for(i = 0; i < tmp; i++)
{
for(j = 0; j <= n; j++)
{
if(dp[i][j] < inf)
{
for(k = 0; k <= n; k++)
{
if(k == 0)
tem = 0;
else tem = (1<<(k-1));
if(!(i&tem))
{
dp[i+tem][k] = min(dp[i+tem][k],dp[i][j]+map[j][k]);//状态为i,经过j的城市，走上j通往k的不归路
}
}
}
}
}

tmp--;
tem = inf;
for(i = 1; i <= n; i++)
{
tem = min(tem,dp[tmp][i]+map[i][0]);//<span style="font-family: Arial, Helvetica, sans-serif;">dp[tmp][i]是已经全部到过，但是终点为i，map[i][0]是快速从I返回到店的捷径</span>

}
printf("%d\n",tem);
}
return 0;
}