[BZOJ1624][Usaco2008 Open] Clear And Present Danger 寻宝之路

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

Time Limit: 5 Sec  Memory Limit: 64 MB Submit: 782  Solved: 512 [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.

floyd

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std; 
char buf[10000000], *ptr = buf - 1;
inline int readint(){
    int n = 0;
    char ch = *++ptr;
    while(ch < '0' || ch > '9') ch = *++ptr;
    while(ch <= '9' && ch >= '0'){
        n = (n << 1) + (n << 3) + ch - '0';
        ch = *++ptr;
    }
    return n;
}
const int maxn = 100 + 10, maxm = 10000 + 10;
int f[maxn][maxn], a[maxm];
int main(){
    fread(buf, sizeof(char), sizeof(buf), stdin);
    int N, M;
    N = readint();
    M = readint();
    for(int i = 1; i <= M; i++) a[i] = readint();
    for(int i = 1; i <= N; i++)
        for(int j = 1; j <= N; j++)
            f[i][j] = readint();
    for(int k = 1; k <= N; k++)
        for(int i = 1; i <= N; i++)
            for(int j = 1; j <= N; j++)
                f[i][j] = min(f[i][j], f[i][k] + f[k][j]);
    long long ans = 0;
    for(int i = 2; i <= M; i++) ans += f[a[i - 1]][a[i]];
    printf("%lld
", ans);
    return 0;
};