Wow! Such Conquering!

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2618    Accepted Submission(s): 790

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

Recommend

liuyiding

求1到每个点的最短时间而且不能超过截止时间。输出每个Doge Planet的所有到达时间的最小总和。

如果超截止时间就输出 -1

先用floyd松弛一下点与点之间的最短距离，然后再dfs暴搜最小的和

剪枝，一、每一步的时间和sumtim必须小于结果ans，

二、到达每一步的时间都必须要保证到达后面点的所有时间都来得及

#include<iostream>
#include<cstdio>
#include<cstring>
#define inf 0x3f3f3f3f
using namespace std;
int n,ans;
int map[35][35],dl[35],maxdl;
int vis[35]; 
void floyd()
{
	for(int k=1;k<=n;k++)
	{
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=n;j++)
			{
				if(map[i][j]>map[i][k]+map[k][j])
				{
					map[i][j]=map[i][k]+map[k][j];
				}
			}
		}
	}
}
void dfs(int pos,int hasnum,int tim,int kk,int sumtim)
{
	if(tim>maxdl) return;
	if(tim>dl[pos]) return;
	if(sumtim>ans) return;//很重要 
	for(int i=2;i<=n;i++)
	{
		if(!vis[i]&&tim+map[pos][i]>dl[i]) return;
	}
	if(hasnum==n)
	{
		if(sumtim<ans) ans=sumtim;
		return;
	}
	for(int i=2;i<=n;i++)
	{
		if(vis[i]) continue;
		vis[i]=1;
		dfs(i,hasnum+1,tim+map[pos][i],kk-1,sumtim+map[pos][i]*kk);//sumtim预处理了很多组数据 
		vis[i]=0;
	}
}
int main()
{
	while(~scanf("%d",&n))
	{
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=n;j++)
			{
				scanf("%d",&map[i][j]);
			}
		}
		maxdl=-1;
		for(int i=2;i<=n;i++)
		{
			scanf("%d",&dl[i]);
			if(maxdl<dl[i]) maxdl=dl[i];
		}
		floyd();
		ans=inf;
		memset(vis,0,sizeof(vis));
		dfs(1,1,0,n-1,0);
		if(ans==inf) printf("-1\n");
		else printf("%d\n",ans);
	}
	return 0;
}