Wow! Such Conquering!

Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2683 Accepted Submission(s): 815

Problem Description

There are n Doge Planets in the Doge Space. The conqueror of Doge Space is Super Doge, who is going to inspect his Doge Army on all Doge Planets. The inspection starts from Doge Planet 1 where DOS (Doge Olympic Statue) was built. It takes Super Doge exactly Txy time to travel from Doge Planet x to Doge Planet y.
With the ambition of conquering other spaces, he would like to visit all Doge Planets as soon as possible. More specifically, he would like to visit the Doge Planet x at the time no later than Deadlinex. He also wants the sum of all arrival time of each Doge Planet to be as small as possible. You can assume it takes so little time to inspect his Doge Army that we can ignore it.

Input

There are multiple test cases. Please process till EOF.
Each test case contains several lines. The first line of each test case contains one integer: n, as mentioned above, the number of Doge Planets. Then follow n lines, each contains n integers, where the y-th integer in the x-th line is Txy . Then follows a single line containing n - 1 integers: Deadline2 to Deadlinen.
All numbers are guaranteed to be non-negative integers smaller than or equal to one million. n is guaranteed to be no less than 3 and no more than 30.

Output

If some Deadlines can not be fulfilled, please output “-1” (which means the Super Doge will say “WOW! So Slow! Such delay! Much Anger! . . . ” , but you do not need to output it), else output the minimum sum of all arrival time to each Doge Planet.

Sample Input

4 0 3 8 6 4 0 7 4 7 5 0 2 6 9 3 0 30 8 30 4 0 2 3 3 2 0 3 3 2 3 0 3 2 3 3 0 2 3 3

Sample Output

36 -1

Hint

Explanation: In case #1: The Super Doge travels to Doge Planet 2 at the time of 8 and to Doge Planet 3 at the time of 12, then to Doge Planet 4 at the time of 16. The minimum sum of all arrival time is 36.

Source

2014西安全国邀请赛

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借鉴：https://blog.csdn.net/tobewhatyouwanttobe/article/details/37575983

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
int n,m,ans,tot,len;
int dist[35][35],time[35],R[35],L[35];
void floyd()
{
    int i,j,k;
    for(k=1;k<=n;k++)
    {
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                dist[i][j]=min(dist[i][j],dist[i][k]+dist[k][j]);
            }
        }
    }
}
void dfs(int u,int cost,int sum,int num)  //深搜就是暴力枚举，毕竟只有30个点，所以暴力下也是可以。
{                                         //额，毕竟是暴力枚举，就不分什么最优，不最优了，直接枚举所有就行了
    for(int i=R[0];i!=0;i=R[i])           //R,L是全局变量。深搜下来的话，有些是被删除了的
    {
        if(cost+dist[u][i]>time[i])
        return;                           //额，直接全部判是否超期限，有一个不行，就不用这种方法，这样的确可以省许多时间
    }                                     //这算是普通的条件剪枝吧。
    if(sum>=ans)                          //A*剪枝。
    return;
    R[L[u]]=R[u];L[R[u]]=L[u];            //删除u，
    if(R[0]==0)                           //当所有点都试过
    {
        ans=min(ans,sum);
        R[L[u]]=L[R[u]]=u;
        return ;
    }
    for(int i=R[0];i!=0;i=R[i])
    {
        dfs(i,cost+dist[u][i],sum+dist[u][i]*num,num-1);  //深搜呈现了一种递进性。
    }                             //sum+dist[u][i]*num是预测出来的最优解。
    R[L[u]]=L[R[u]]=u;            //枚举回溯吧
    return;
}
int main()
{
   int i,j;
   while(~scanf("%d",&n))
   {
       for(i=1;i<=n;i++)
       {
           for(j=1;j<=n;j++)
           {
               scanf("%d",&dist[i][j]);
           }
       }
       for(i=2;i<=n;i++)
       {
           scanf("%d",&time[i]);
       }
       floyd();
       for(i=0;i<=n;i++)  //双向链表的建立
       {
           R[i]=i+1; L[i]=i-1;
       }
       R[n]=0; L[0]=n;
       ans=INF;
       dfs(1,0,0,n-1);   //num一开始从n-1开始的，好吧
       if(ans==INF)
       ans=-1;
       printf("%d\n",ans);
   }
}

HDU-4848 Wow! Such Conquering!(A*)

Wow! Such Conquering!

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