lightoj 1049 - One Way Roads 【DFS】

题目链接：lightoj 1049 - One Way Roads

1049 - One Way Roads
PDF (English) Statistics Forum
Time Limit: 0.5 second(s) Memory Limit: 32 MB
Nowadays the one-way traffic is introduced all over the world in order to improve driving safety and reduce traffic jams. The government of Dhaka Division decided to keep up with new trends. Formerly all n cities of Dhaka were connected by n two-way roads in the ring, i.e. each city was connected directly to exactly two other cities, and from each city it was possible to get to any other city. Government of Dhaka introduced one-way traffic on all n roads, but it soon became clear that it’s impossible to get from some of the cities to some others. Now for each road is known in which direction the traffic is directed at it, and the cost of redirecting the traffic. What is the smallest amount of money the government should spend on the redirecting of roads so that from every city you can get to any other?

Input
Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case starts with a blank line and an integer n (3 ≤ n ≤ 100) denoting the number of cities (and roads). Next n lines contain description of roads. Each road is described by three integers ai, bi, ci (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ ci ≤ 100) - road is directed from city ai to city bi, redirecting the traffic costs ci.

Output
For each case of input you have to print the case number and the smallest amount of money the government should spend on the redirecting of roads so that from every city you can get to any other.

Sample Input
Output for Sample Input
4

3
1 3 1
1 2 1
3 2 1

3
1 3 1
1 2 5
3 2 1

6
1 5 4
5 3 8
2 4 15
1 6 16
2 3 23
4 6 42

4
1 2 9
2 3 8
3 4 7
4 1 5
Case 1: 1
Case 2: 2
Case 3: 39
Case 4: 0

题意：n个点，每个点的入度、出度确定为1。给出n条有向边和反向该边的代价。问你连通这n个点的最小代价。

思路：题目给出的图，去掉一条边一定是一棵树，那么我们直接遍历一次，找到1-n的路径，一定是唯一的。按照这条路径统计花费就可以了。

AC代码：

#include <cstdio>#include <cstring>#include <vector>#define CLR(a, b) memset(a, (b), sizeof(a))using namespace std;vector<int> G[110];bool vis[110];int n;int Map[110][110];int rec[110], top;void DFS(int u, int sum) {    vis[u] = true;    if(sum == n) {        return ;    }    for(int i = 0; i < G[u].size(); i++) {        int v = G[u][i];        if(vis[v]) continue;        rec[top++] = v; DFS(v, sum + 1);    }}int main(){    int t, kcase = 1;    scanf("%d", &t);    while(t--) {        scanf("%d", &n);        for(int i = 1; i <= n; i++) {            G[i].clear(); vis[i] = false;        }        CLR(Map, 0);        for(int i = 1; i <= n; i++) {            int u, v, w;            scanf("%d%d%d", &u, &v, &w);            G[u].push_back(v);            G[v].push_back(u);            Map[u][v] = w;        }        top = 0; rec[top++] = 1; DFS(1, 1);        int ans1 = 0;        for(int i = 1; i < top; i++) {            int u = rec[i-1]; int v = rec[i];            if(Map[u][v] == 0) {                ans1 += Map[v][u];            }        }        if(Map[rec[top-1]][1] == 0) ans1 += Map[1][rec[top-1]];        int ans2 = 0;        for(int i = top-1; i >= 1; i--) {            int u = rec[i]; int v = rec[i-1];            if(Map[u][v] == 0) {                ans2 += Map[v][u];            }        }        if(Map[1][rec[top-1]] == 0) ans2 += Map[rec[top-1]][1];        printf("Case %d: %d\n", kcase++, min(ans1, ans2));    }    return 0;}