题目链接 https://vjudge.net/problem/HihoCoder-1689

Sample Input

5 8
A1 < A3
A3 < A2
A3 < A4
A5 < A2
A1 < A4
A1 < A2
A5 < A1
A5 < A3

Sample Output

7

【思路】

图论基础+二分法 如果a<b就加一条有向边a->b如果a==b就加两条有向边a->b和b->a，然后用一个变量index记录这条边是输入中从上往下数的第几条边,采用二分法判断前x条边内否判断出最终的大小关系，在二分的时候要用两次dfs，一次是看能否从起点走到终点，另一次是从终点到起点，只要有一次连通着就可以，dfs搜索的时候所经过的边不能超过mid,最后就能出答案了。输入格式比较坑，我也是看另一个人的题解才知道的。

#include<bits/stdc++.h>
using namespace std;const int maxn = 1e5 + 50;
const int maxm = 1e6 + 50;int n, m, from, to;
int le, ri, mid;
char op[2];struct node {int to, index;node(int t, int in) :to(t), index(in) {}
};bool used[maxn];
vector<node> g[maxn];void init() {for (int i = 0; i < maxn; ++i) g[i].clear();
}bool dfs1(int u) {if (u == 1) return 1;used[u] = 1;for (int i = 0; i < g[u].size(); ++i) {node e = g[u][i];if (e.index > mid) continue;if (used[e.to]) continue;if (dfs1(e.to)) return 1;}return 0;
}bool dfs2(int u) {if (u == n) return 1;used[u] = 1;for (int i = 0; i < g[u].size(); ++i) {node e = g[u][i];if (e.index > mid) continue;if (used[e.to]) continue;if (dfs2(e.to)) return 1;}return 0;
}int main() {init();scanf("%d%d ", &n, &m);for (int i = 1; i <= m; ++i) {getchar();scanf("%d ", &from);scanf("%s ", op);getchar();scanf("%d ", &to);if (from == to) continue;if ('<' == op[0]) {g[from].push_back(node(to, i));}else if ('=' == op[0]) {g[from].push_back(node(to, i));g[to].push_back(node(from, i));}}if (n == 1) {printf("0\n");return 0;} int ans = -1;le = 0, ri = m;while (le + 1 < ri) {mid = (le + ri) >> 1;memset(used, 0, sizeof(used));bool ok = dfs1(n);memset(used, 0, sizeof(used));if (ok || dfs2(1)) { ans = mid; ri = mid; }else { le = mid; }}printf("%d\n", ans);return 0;
}