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 tovisit a sequence of 4 islands in order: island 1, island 2, island1 again, and finally island 3. The danger ratings of the paths aregiven: the paths (1, 2); (2, 3); (3, 1) and the reverse paths havedanger ratings of 5, 2, and 1, respectively.
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 tovisit a sequence of 4 islands in order: island 1, island 2, island1 again, and finally island 3. The danger ratings of the paths aregiven: the paths (1, 2); (2, 3); (3, 1) and the reverse paths havedanger 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 inthe 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 pathbetween islands 1 and 2 because it has a large danger rating.
OUTPUT DETAILS:
He can get the treasure with a total danger of 7 by traveling inthe 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 pathbetween islands 1 and 2 because it has a large danger rating.
看到是floyd感觉好久没有写过了就水一发。
结果maxm写成maxn。智障错误第二次：手打堆
//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;
const int maxn=100+10,maxm=1e4+10;
int n,m,a[maxm];//
long long dis[maxn][maxn],ans;
long long aa;char cc;
long long read() {
aa=0;cc=getchar();
while(cc<'0'||cc>'9') cc=getchar();
while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
return aa;
}
int main() {
n=read();m=read();a[0]=1;
for(int i=1;i<=m;++i) a[i]=read();a[m+1]=n;
for(int i=1;i<=n;++i) for(int j=1;j<=n;++j) dis[i][j]=read();
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]);
for(int i=0;i<=m;++i) ans+=dis[a[i]][a[i+1]];
cout<<ans;
return 0;
}