BZOJ 1624: [Usaco2008 Open] Clear And Present Danger 寻宝之路 (Floyd)

1624: [Usaco2008 Open] Clear And Present Danger 寻宝之路

Time Limit: 5 Sec  Memory Limit: 64 MB Submit: 507  Solved: 345 [Submit][Status][Discuss]

Description

    农夫约翰正驾驶一条小艇在牛勒比海上航行．

    海上有N(1≤N≤100)个岛屿，用1到N编号．约翰从1号小岛出发，最后到达N号小岛．一

张藏宝图上说，如果他的路程上经过的小岛依次出现了Ai，A2，…，AM(2≤M≤10000)这样的序列（不一定相邻），那他最终就能找到古老的宝藏．  但是，由于牛勒比海有海盗出没．约翰知道任意两个岛屿之间的航线上海盗出没的概率，他用一个危险指数Dij(0≤Dij≤100000)来描述．他希望他的寻宝活动经过的航线危险指数之和最小．那么，在找到宝藏的前提下，这个最小的危险指数是多少呢？

Input

    第1行输入N和M，之后M行一行一个整数表示A序列，之后输入一个NxN的方阵，表示两两岛屿之间航线的危险指数．数据保证Dij=Dji，Dii=0．

Output

    最小的危险指数和．

Sample Input

3 4 1 2 1 3 0 5 1 5 0 2 1 2 0

INPUT DETAILS:

There are 3 islands and the treasure map requires Farmer John to visit a sequence of 4 islands in order: island 1, island 2, island 1 again, and finally island 3. The danger ratings of the paths are given: the paths (1, 2); (2, 3); (3, 1) and the reverse paths have danger ratings of 5, 2, and 1, respectively.

Sample Output

7

OUTPUT DETAILS:

He can get the treasure with a total danger of 7 by traveling in the sequence of islands 1, 3, 2, 3, 1, and 3. The cow map's requirement (1, 2, 1, and 3) is satisfied by this route. We avoid the path between islands 1 and 2 because it has a large danger rating.

HINT

Source

Silver

思路：可以把危险指数看成距离，于是floyd一下就好咯

/* ***********************************************
Author :111qqz
Created Time :2016年04月04日 星期一 02时54分18秒
File Name :code/bzoj/1624.cpp
************************************************ */

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair

using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N=105;
int d[N][N];
int n,m;
int a[10005];
int main()
{
	#ifndef  ONLINE_JUDGE 
	freopen("code/in.txt","r",stdin);
  #endif

	ios::sync_with_stdio(false);
	cin>>n>>m;
	a[0]=1;
	a[m+1]=n;
	for ( int i = 1 ; i <= m ; i++) cin>>a[i];
    
	for ( int i = 1 ; i <= n ; i++) 
	    for ( int j = 1 ; j <= n ; j++) cin>>d[i][j];
	
	for ( int k = 1 ; k <= n ; k++)
	    for ( int i = 1 ; i <= n ; i++)
		for ( int j = 1 ; j <= n ; j++)
		    d[i][j] = min(d[i][j],d[i][k]+d[k][j]);

	int ans = 0 ;
	for ( int i = 1 ; i <= m+1 ; i++)
	    ans +=d[a[i-1]][a[i]];

	cout<<ans<<endl;
    
  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}