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: Fire Station

A city is served by a number of fire stations. Some residents have complained that the distance from their houses to the nearest station is too far, so a new station is to be built. You are to choose the location of the fire station so as to reduce the distance to the nearest station from the houses of the disgruntled residents.

The city has up to 500 intersections, connected by road segments of various lengths. No more than 20 road segments intersect at a given intersection. The location of houses and firestations alike are considered to be at intersections (the travel distance from the intersection to the actual building can be discounted). Furthermore, we assume that there is at least one house associated with every intersection. There may be more than one firestation per intersection.

The Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

The first line of input contains two positive integers: f,the number of existing fire stations (f <= 100) and i, the number of intersections (i <= 500). The intersections are numbered from 1 to i consecutively. f lines follow; each contains the intersection number at which an existing fire station is found. A number of lines follow, each containing three positive integers: the number of an intersection, the number of a different intersection, and the length of the road segment connecting the intersections. All road segments are two-way (at least as far as fire engines are concerned), and there will exist a route between any pair of intersections.

The Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

You are to output a single integer: the lowest intersection number at which a new fire station should be built so as to minimize the maximum distance from any intersection to the nearest fire station.

Sample Input

1

1 6
2
1 2 10
2 3 10
3 4 10
4 5 10
5 6 10
6 1 10

Output for Sample Input

5

---

先講解一下題目內容，平面上有 n 個點，其中 f 個點是消防站的位置，居民要到達消防站，現在請多設立一個消防站，使得這些居民到達消防站的最大值最小化。

很明顯地大部份人都使用 dijsktra, 但是在邊很少, 而且重複要對居民去找到最鄰近的消防站, 用 spfa 是比較好寫的, 而且效率比較好。

#include <stdio.h>
#include <queue>
#define oo 0xffffff
using namespace std;
int fire[505];
int g[505][505][2], gt[505];
int dist[505], f, n;
void spfa(int nd) {
    int i, j, tn, mn;
    char used[505] = {};
    queue<int> Q;
    for(i = 0; i <= n; i++)
        used[i] = 0;
    dist[nd] = 0;
    Q.push(nd);
    while(!Q.empty()) {
        tn = Q.front();
        Q.pop();
        used[tn] = 0;
        for(j = 0; j < gt[tn]; j++) {
            if(dist[g[tn][j][0]] > dist[tn]+g[tn][j][1]) {
                dist[g[tn][j][0]] = dist[tn]+g[tn][j][1];
                if(!used[g[tn][j][0]])
                    Q.push(g[tn][j][0]), used[g[tn][j][0]] = 1;
            }
        }
    }
}
int main() {
    int t, i, j;
    int a, b, c;
    char in[505];
    scanf("%d", &t);
    while(t--) {
        scanf("%d %d", &f, &n);
        for(i = 0; i < f; i++)
            scanf("%d", &fire[i]);
        for(i = 0; i <= n; i++)
            gt[i] = 0, dist[i] = oo;
        gets(in);
        while(gets(in)) {
            if(!in[0])  break;
            sscanf(in, "%d %d %d", &a, &b, &c);
            g[a][gt[a]][0] = b, g[a][gt[a]++][1] = c;
            g[b][gt[b]][0] = a, g[b][gt[b]++][1] = c;
        }
        for(i = 0; i < f; i++)
            spfa(fire[i]);
        int o[505];
        for(i = 1; i <= n; i++)
            o[i] = dist[i];
        int ans = oo, anum;
        for(i = 1; i <= n; i++) {
            spfa(i);
            int mx = -oo;
            for(j = 1; j <= n; j++) {
                if(mx < dist[j])
                    mx = dist[j];
            }
            if(ans > mx)    ans = mx, anum = i;
            for(j = 1; j <= n; j++)
                dist[j] = o[j];
        }
        printf("%d\n", anum);
        if(t)   puts("");
    }
    return 0;
}