Codeforce336D.Dima and Trap Graph(二分右端点+dfs验证)

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题目传送门：http://codeforces.com/problemset/problem/366/D

题意：有n个点m条无向边。初始你需要选择一个整数x，走第i条边的限制为Li <= x <= Ri，假设1-n的一条路径上可以选择的整数x有ans个，问你最大的ans。

分析：因为答案肯定是这些边中最优左端点到某个右端点的，所以就成了在已知左端点的情况下，维护最大的右端点。二分右端点，dfs验证是否可行即可。

#include<bits/stdc++.h>#define PI acos(-1.0)#define eps 1e-8#define ll long long#define MEM(a, b) memset(a, b, sizeof(a))#define pb push_back#define mp make_pair#define MII map<int,int>::iterator#define MLL map<LL,LL>::iterator#define pii pair<int,int>#define si set<int>::iterator#define sl set<LL>::iterator#define bug printf("bug-------bug-------bug\n")using namespace std;const int maxn = 1005;int Time, head[maxn], vis[maxn], eid, ans, n, m;bool success;set<int> st;struct Edge{    int to, x, y, nxt;}edge[6010];void addedge(int u, int v, int l, int r){    edge[eid].to = v;    edge[eid].x = l;    edge[eid].y = r;    edge[eid].nxt = head[u];    head[u] = eid++;}void dfs(int u, int l, int r){    if(success)        return;    if(u == n)    {        success = true;        return;    }    for(int i = head[u]; i != -1; i = edge[i].nxt)    {        int v = edge[i].to;        if(vis[v] == Time || edge[i].x > l || edge[i].y < r)            continue;        vis[v] = Time;        dfs(v, l, r);    }}void solve(){    ans = 0;    for(si i = st.begin(); i != st.end(); i++)    {        int now = *i;        int left = now, right = 1e6+5, mid;        while(left < right)        {            Time++;            success = false;            vis[1] = Time;            mid = (left + right) / 2;            dfs(1, now, mid);            if(success)                left = mid + 1;            else                right = mid;        }        ans = max(ans, left - now);    }}int main(){    Time = 1;    memset(vis, 0, sizeof(vis));    while(cin >> n >> m)    {        eid = 0;        memset(head, -1, sizeof(head));        st.clear();        while(m--)        {            int u, v, l, r;            cin >> u >> v >> l >> r;            addedge(u, v, l, r);            addedge(v, u, l, r);            st.insert(l);            st.insert(r);        }        solve();        if(ans == 0)            printf("Nice work, Dima!\n");        else            printf("%d\n",ans);    }    return 0;}

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