POJ3311:Hie with the Pie(floyd+状态压缩DP)

Description

ThePizazzPizzeriapridesitselfindeliveringpizzastoitscustomersasfastaspossible.Unfortunately,duetocutbacks,theycanaffordtohireonlyonedrivertodothedeliveries.Hewillwaitfor1ormore(upto10)orderstobeprocessedbefore
hestartsanydeliveries.Needlesstosay,hewouldliketotaketheshortestrouteindeliveringthesegoodiesandreturningtothepizzeria,evenifitmeanspassingthesamelocation(s)orthepizzeriamorethanonceontheway.Hehascommissionedyou
towriteaprogramtohelphim.

Input

Inputwillconsistofmultipletestcases.Thefirstlinewillcontainasingleinteger
nindicatingthenumberoforderstodeliver,where1≤n≤10.Afterthiswillbe
n+1lineseachcontainingn+1integersindicatingthetimestotravelbetweenthepizzeria(numbered0)andthe
nlocations(numbers1ton).Thejthvalueonthe
ithlineindicatesthetimetogodirectlyfromlocationitolocation
jwithoutvisitinganyotherlocationsalongtheway.Notethattheremaybequickerwaystogofrom
itojviaotherlocations,duetodifferentspeedlimits,trafficlights,etc.Also,thetimevaluesmaynotbesymmetric,i.e.,thetimetogodirectlyfromlocation
itojmaynotbethesameasthetimetogodirectlyfromlocation
jtoi.Aninputvalueofn=0willterminateinput.

Output

Foreachtestcase,youshouldoutputasinglenumberindicatingtheminimumtimetodeliverallofthepizzasandreturntothepizzeria.

SampleInput

3
011010
1012
101010
102100
0

SampleOutput

8

一个送外卖的人，要将外卖全部送去所有地点再回到店离，求最短路

首先用floyd求出最短路，然后再进行DP即可

dp[i][j]表示i这个状态下，目标是j的最短路，i是用二进制表示每个地点是否去过

#include<stdio.h>
#include<string.h>
#include<algorithm>
usingnamespacestd;

intmap[20][20],dis[20][20],dp[1<<11][20];

intmain()
{
intn,i,j,k;
while(~scanf("%d",&n),n)
{
for(i=0;i<=n;i++)
for(j=0;j<=n;j++)
{
scanf("%d",&map[i][j]);
dis[i][j]=map[i][j];
}
for(j=0;j<=n;j++)
for(i=0;i<=n;i++)
for(k=0;k<=n;k++)
if(dis[i][j]>dis[i][k]+map[k][j])
dis[i][j]=dis[i][k]+map[k][j];
memset(dp,-1,sizeof(dp));
dp[1][0]=0;
for(i=1;i<1<<(n+1);i++)
{
i=i|1;
for(j=0;j<=n;j++)
{
if(dp[i][j]!=-1)
{
for(k=0;k<=n;k++)
{
if(j!=k&&(dp[(1<<k)|i][k]==-1||dp[(1<<k)|i][k]>dp[i][j]+dis[j][k]))
dp[(1<<k)|i][k]=dp[i][j]+dis[j][k];
}
}
}
}
printf("%d\n",dp[(1<<(n+1))-1][0]);
}

return0;
}