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  • Codeforces 336D Dima and Trap Graph 并查集

    Dima and Trap Graph

    枚举区间的左端点, 然后那些左端点比枚举的左端点小的都按右端点排序然后并查集去check

    #include<bits/stdc++.h>
    #define LL long long
    #define fi first
    #define se second
    #define mk make_pair
    #define PLL pair<LL, LL>
    #define PLI pair<LL, int>
    #define PII pair<int, int>
    #define SZ(x) ((int)x.size())
    #define ull unsigned long long
    using namespace std;
    
    const int N = 3e3 + 7;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 1e9 + 7;
    const double eps = 1e-8;
    
    int n, m, tot, ans;
    int fa[N];
    struct Edge {
        bool operator < (const Edge& rhs) const {
            return l < rhs.l;
        }
        int a, b, l, r;
    } e[N], tmp[N];
    
    int getRoot(int x) {
        return fa[x] == x ? x : fa[x] = getRoot(fa[x]);
    }
    
    int Merge(int id) {
        int u = tmp[id].a, v = tmp[id].b;
        fa[getRoot(u)] = getRoot(v);
        if(getRoot(1) == getRoot(n)) return tmp[id].r;
        else return -1;
    }
    int main() {
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= m; i++)
            scanf("%d%d%d%d", &e[i].a, &e[i].b, &e[i].l, &e[i].r);
        sort(e + 1, e + 1 + m);
        for(int i = 1; i <= m; i++) {
            for(int j = 1; j <= n; j++) fa[j] = j;
            tmp[++tot] = e[i];
            int ret = Merge(tot);
            for(int j = 1; j < tot && ret == -1; j++) ret = Merge(j);
            if(~ret) ans = max(ans, min(e[i].r, ret) - e[i].l + 1);
            for(int j = tot; j > 1; j--)
                if(tmp[j].r > tmp[j - 1].r) swap(tmp[j], tmp[j - 1]);
        }
        if(ans) printf("%d
    ", ans);
        else puts("Nice work, Dima!");
        return 0;
    }
    
    /*
    */
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  • 原文地址:https://www.cnblogs.com/CJLHY/p/10415559.html
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