Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem A

Airport Setup

Time Limit: 3 Second

There are N cities in the kingdom ‘Nameless’ and among them K cities have airport. There the airline companies are willing to setup their flight directly between M pairs of the cities. They think that it will not be profitable for them to run flights between other pairs of cities. Note that if there are airports in both of the cities of any of their M pairs only then they will run flight between that pair otherwise not. Because of this policy of the airline companies and also because all the cities don’t have airport many people can’t travel between their desired cities. So some people of the country have demanded that the king should do something so that they can travel to their destinations. There are Q demands which came to the king. Every demand is a pair (x, y) which means some people want a route between city x and city y. The king really wants to satisfy the demands but he does not want to setup too many airports. So for every demand (x, y) the king wants to know the minimum number of airports he needs to setup to establish a route between city x and city y.

Input:

The first line of the input contains X the number of test cases which is at most 10. Each case starts with three numbers N ( 1 ≤ N ≤ 2,000), M ( 0 ≤ M ≤ 10,000) and K ( 1 ≤ K ≤ N). Here N is the number of cities, M is the number of direct air routes the airlines companies are willing to setup and K is the number of cities which has airport. Then there will be a line containing K integers which denote the name of the cities which have airport (city name ranges from 1 to N). Each city is separated by a single space. Then there will be M lines each of which contains one pair “a b”. It means that the airline companies are willing to run their flights between city a and b. The next line contains Q ( 1 ≤ Q ≤ 50 ) which is the number of demands came to the king. Then there will be Q lines each containing a pair “x y” which means that the king should build minimum airport to establish a route between city x and city y. There is a blank line between every consecutive test case.

Output:

For every test case you have to output the case number first (See the sample). Then you have to output Q lines each containing the minimum number of airports the king has to setup to satisfy that demand. If it’s impossible to satisfy that demand then output -1. You have to output a blank line after each test case.

Problemsetter: Tasnim Imran Sunny

Special Thanks To: Md. Mahbubul Hasan

#include <stdio.h>
#include <queue>
#include <vector>
using namespace std;
vector<int> g[2005];
int airport[2005];
struct E {
    int node, v;
    E(int a, int b):
        node(a), v(b) {}
    bool operator<(const E &A) const {
        return v > A.v;
    }
};
void dijkstra(int st, int ed, int n) {
    if(st == ed) {
        puts("0");
        return;
    }
    priority_queue<E> pQ;
    E tn(0,0);
    int dist[n+1], i;
#define oo 0xffffff
    for(i = 0; i <= n; i++)
        dist[i] = oo;
    dist[st] = (airport[st] == 0);
    pQ.push(E(st, 0));
    while(!pQ.empty()) {
        tn = pQ.top(), pQ.pop();
        if(tn.v >= dist[ed])  continue;
        for(vector<int>::iterator it = g[tn.node].begin();
            it != g[tn.node].end(); it++) {
            if(airport[*it] == 1) {
                if(dist[*it] > dist[tn.node]) {
                    dist[*it] = dist[tn.node];
                    pQ.push(E(*it, dist[*it]));
                }
            } else {
                if(dist[*it] > dist[tn.node]+1) {
                    dist[*it] = dist[tn.node]+1;
                    pQ.push(E(*it, dist[*it]));
                }
            }
        }
    }
    if(dist[ed] == oo)
        puts("-1");
    else
        printf("%d\n", dist[ed]);
}
int main() {
    int testcase, cases = 0;
    int i, j, k, n, m, q;
    int x, y;
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d %d %d", &n, &m, &k);
        for(i = 1; i <= n; i++)
            g[i].clear(), airport[i] = 0;
        while(k--) {
            scanf("%d", &x);
            airport[x] = 1;
        }
        while(m--) {
            scanf("%d %d", &x, &y);
            g[x].push_back(y);
            g[y].push_back(x);
        }
        scanf("%d", &q);
        printf("Case %d:\n", ++cases);
        while(q--) {
            scanf("%d %d", &x, &y);
            dijkstra(x, y, n);
        }
        puts("");
    }
    return 0;
}