题目链接如下：

Online Judge

脑雾严重，这道题一开始我想的方向有问题.....后来看了别人的题解才写出来的.....

用的是欧拉路径的充要条件；以及数连通块。需要加的高速路数目 = 连通块个数 - 1 + sum（每个连通块中连成欧拉路径需要加的高速路数目）。

#include <cstdio>
#include <algorithm>
// #define debugint V, E, T, a, b, tot, odd, kase = 0;
int arc[1001][1001];
bool vis[1001];
int cnt[1001];void dfs(int k){vis[k] = true;if (cnt[k] % 2){odd++;}for (int i = 1; i <= V; ++i){if (!vis[i] && arc[i][k]){dfs(i);}}
}int main(){#ifdef debugfreopen("0.txt", "r", stdin);freopen("1.txt", "w", stdout);#endifwhile (scanf("%d %d %d", &V, &E, &T) == 3 && (V || E || T)){std::fill(vis, vis + V + 1, true);std::fill(cnt, cnt + V + 1, 0);tot = E - 1;for (int i = 0; i <= V; ++i){for (int j = 0; j <= V; ++j){arc[i][j] = arc[j][i] = 0;}}for (int i = 0; i < E; ++i){scanf("%d %d", &a, &b);arc[a][b] = arc[b][a] = 1;vis[a] = vis[b] = false;cnt[a]++;cnt[b]++;}for (int i = 1; i <= V; ++i){if (!vis[i]){odd = 0;dfs(i);if (odd > 2){tot += (odd - 2) / 2;}tot++;}}printf("Case %d: %d\n", ++kase, E == 0 ? 0 : tot * T);}#ifdef debugfclose(stdin);fclose(stdout);#endifreturn 0;
}