CROC 2016 - Elimination Round (Rated Unofficial Edition) D. Robot Rapping Results Report 拓扑排序+二分(示例

Posted 2020-07-15

题目链接：

http://www.codeforces.com/contest/655/problem/D

题意：

题目是要求前k个场次就能确定唯一的拓扑序，求满足条件的最小k。

题解：

二分k的取值，做拓扑排序的时候只要每次只有一个元素没有前驱就可以唯一了。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<utility>
using namespace std;

const int maxn = 101010;
int n, m;

struct Edge {
    int v, ne;
    Edge(int v, int ne) :v(v), ne(ne) {}
    Edge() {}
}egs[maxn*2];

int head[maxn], tot;

void addEdge(int u, int v) {
    egs[tot] = Edge(v, head[u]);
    head[u] = tot++;
}

int ind[maxn];
bool ok(int m) {
    memset(ind, 0, sizeof(ind));
    for (int i = 0; i <= m; i++) {
        ind[egs[i].v]++;
    }
    queue<int> Q;
    for (int i = 0; i < n; i++) {
        if (ind[i] == 0) {
            Q.push(i);
        }
    }
    while (!Q.empty()) {
        if (Q.size() > 1) return false;
        int u = Q.front(); Q.pop();
        int p = head[u];
        while (p != -1) {
            Edge& e = egs[p];
            if (p <= m) {
                ind[e.v]--;
                if (ind[e.v] == 0) {
                    Q.push(e.v);
                }
            }
            p = e.ne;
        }
    }
    return true;
}

void init() {
    memset(head, -1, sizeof(head));
    tot = 0;
}

int main() {
    while (scanf("%d%d", &n, &m) == 2 && n) {
        init();
        for (int i = 0; i < m; i++) {
            int u, v;
            scanf("%d%d", &u, &v); u--, v--;
            addEdge(u, v);
        }
        int l = -1, r = tot-1;
        while (l + 1 < r) {
            int mid = l + (r - l) / 2;
            if (!ok(mid)) l = mid;
            else r = mid;
        }
        //printf("l:%d\n", l);
        if (!ok(r)) printf("-1\n");
        else printf("%d\n", r + 1);
    }
    return 0;
}