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 B: Jogging Trails

Gord is training for a marathon. Behind his house is a park with a large network of jogging trails connecting water stations. Gord wants to find the shortest jogging route that travels along every trail at least once.

Input consists of several test cases. The first line of input for each case contains two positive integers: n <= 15, the number of water stations, and m < 1000, the number of trails. For each trail, there is one subsequent line of input containing three positive integers: the first two, between 1 and n, indicating the water stations at the end points of the trail; the third indicates the length of the trail, in cubits. There may be more than one trail between any two stations; each different trail is given only once in the input; each trail can be travelled in either direction. It is possible to reach any trail from any other trail by visiting a sequence of water stations connected by trails. Gord's route may start at any water station, and must end at the same station. A single line containing 0 follows the last test case.

For each case, there should be one line of output giving the length of Gord's jogging route.

Sample Input

4 5
1 2 3
2 3 4
3 4 5
1 4 10
1 3 12
0

Output for Sample Input

41

---

利用尤拉環道的概念, 每個節點都必須有偶數個 deg,
在此題中, 如果產生奇數 deg 的點, 則將其兩兩匹配, 使其圖變成尤拉環道,
其匹配的成本越小越好

#include <stdio.h>
#include <string.h>
#define min(x, y) ((x) < (y) ? (x) : (y))
int map[16][16], odd[16];
void floyd(int n) {
    int i, j, k;
    for(k = 1; k <= n; k++)
        for(i = 1; i <= n; i++)
            for(j = 1; j <= n; j++)
                map[i][j] = min(map[i][j], map[i][k]+map[k][j]);
}
int dp[1<<16];
int build(int pN, int ot) {
    if(pN == 0)
        return 0;
    if(dp[pN] != -1)
        return dp[pN];
    int i, j, tmp;
    dp[pN] = 0xfffffff;
    for(i = 0; i < ot; i++) {
        if(pN&(1<<i)) {
            for(j = i+1; j < ot; j++) {
                if(pN&(1<<j)) {
                    tmp = build(pN-(1<<i)-(1<<j), ot);
                    dp[pN] = min(dp[pN], tmp+map[odd[i]][odd[j]]);
                }
            }
            break;
        }
    }
    return dp[pN];
}
int main() {
    int n, m, x, y, w;
    while(scanf("%d", &n) == 1 && n) {
        scanf("%d", &m);
        memset(map, 63, sizeof(map));
        memset(dp, -1, sizeof(dp));
        int sum = 0, deg[16] = {};
        while(m--) {
            scanf("%d %d %d", &x, &y, &w);
            map[x][y] = min(map[x][y], w);
            map[y][x] = min(map[y][x], w);
            deg[x]++, deg[y]++;
            sum += w;
        }
        floyd(n);
        int ot = 0;
        for(int i = 1; i <= n; i++)
            if(deg[i]&1)
                odd[ot++] = i;
        printf("%d\n", sum+build((1<<ot)-1, ot));
    }
    return 0;
}