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题意：给定n个城市，其中有k个有仓库，问你在其他n-k个城市离仓库的最短距离是多少。
析：很容易想到暴力，并且要想最短，那么肯定是某一个仓库和某一个城市直接相连，这才是最优，所以只要枚举仓库，找第一个城市，然后更新答案即可。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <stack>
using namespace std ;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 100000000000000000;
const double PI = acos(-1.0);
const double eps = 1e-10;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
vector<int> G[maxn];
vector<int> w[maxn];
bool a[maxn];
int main(){
int k;
scanf("%d %d %d", &n, &m, &k);
memset(a, 0, sizeof(a));
for(int i = 0; i < m; ++i){
int x, y, v;
scanf("%d %d %d", &x, &y, &v);
G[x].push_back(y);
G[y].push_back(x);
w[x].push_back(v);
w[y].push_back(v);
}
for(int i = 0; i < k; ++i){
int x;
scanf("%d", &x);
a[x] = true;
}
if(k == 0){ printf("-1\n"); return 0; }
LL ans = LNF;
for(int i = 1; i <= n; ++i){
if(a[i]){
for(int j = 0; j < G[i].size(); ++j){
int u = G[i][j];
if(!a[u]) ans = Min(ans, (LL)w[i][j]);
}
}
}
if(ans == LNF) printf("-1\n");
else printf("%I64d\n", ans);
return 0;
}
CodeForces 707B Bakery (水题，暴力，贪心)
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原文地址：http://www.cnblogs.com/dwtfukgv/p/5793075.html
