Wow! Such Conquering!

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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
T_xy 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 Deadline_x. 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 T_xy . Then
follows a single line containing n - 1 integers: Deadline₂ to Deadline_n.

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

题意：n个点，要求从1点开始跑，遍历所有点，要求到达每个点的时间<=time[i],求到达每个点时间的总和最小。

爆搜+最优性剪枝+可行性剪枝（参考别人的）
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#define L long long
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=32;
int dis[maxn][maxn];
int n,tim[maxn];
void Floyd()
{
	for(int k=1;k<=n;k++)
		for(int i=1;i<=n;i++)
		   for(int j=1;j<=n;j++)
		   dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
}
int vis[maxn],ans;
void dfs(int src,int cur_time,int sum_time,int num)
{
	if(num==n)
	{
		ans=min(ans,sum_time);
		return ;
	}
	//最优性剪枝：当前总时间+当前时间*剩余点数（因为往后到达的点的时间总是小于等于当前时间的）>=当前最优解 剪掉
	if(sum_time+cur_time*(n-num)>=ans)return ;
	//可行性剪枝：如果当前时间有超Deadtime的，剪掉
	for(int i=2;i<=n;i++)
		if(!vis[i]&&cur_time>tim[i])
		return ;
	for(int i=2;i<=n;i++)
	{
		if(src!=i&&!vis[i]&&cur_time+dis[src][i]<=tim[i])
		{
			vis[i]=1;
			dfs(i,cur_time+dis[src][i],sum_time+cur_time+dis[src][i],num+1);
			vis[i]=0;
		}
	}

}
int main()
{
	while(~scanf("%d",&n))
	{
		memset(vis,0,sizeof(vis));
		for(int i=1;i<=n;i++)
			for(int j=1;j<=n;j++)
			scanf("%d",&dis[i][j]);
		tim[1]=INF;
		for(int i=2;i<=n;i++)
			scanf("%d",&tim[i]);
		Floyd();
		vis[1]=1;
		ans=INF;
		dfs(1,0,0,1);
		if(ans==INF)ans=-1;
		printf("%d\n",ans);
	}
	return 0;
}