P3275 [SCOI2011] 糖果

合集 - 图论(79)

1.F. Chat Screenshots2024-02-192.P1656 炸铁路2024-02-223.P1137 旅行计划2024-02-244.P2835 刻录光盘2024-02-245.P1197 [JSOI2008] 星球大战2024-02-246.P3388 【模板】割点（割顶）2024-02-247.P8435 【模板】点双连通分量2024-02-258.P8436 【模板】边双连通分量2024-02-259.P2860 [USACO06JAN] Redundant Paths G2024-02-2510.P1653 [USACO04DEC] Cow Ski Area G2024-02-2611.P3047 [USACO12FEB] Nearby Cows G2024-03-0412.P1894 [USACO4.2] 完美的牛栏The Perfect Stall2024-03-0613.P1550 [USACO08OCT] Watering Hole G2024-03-0614.P2330 [SCOI2005] 繁忙的都市2024-03-0615.P1525 [NOIP2010 提高组] 关押罪犯2024-03-0716.P1379 八数码难题2024-03-0717.P2746 [USACO5.3] 校园网Network of Schools2024-03-0818.P6121 [USACO16OPEN] Closing the Farm G2024-03-0819.P2341 [USACO03FALL / HAOI2006] 受欢迎的牛 G2024-03-0820.P2055 [ZJOI2009] 假期的宿舍2024-03-0821.P5905 【模板】全源最短路（Johnson）2024-03-1122.F. Microcycle2024-03-1223.G. Path Prefixes2024-03-1424.G. Rudolf and Subway2024-03-1525.C. Ehab and Path-etic MEXs2024-03-1726.A. String Transformation 12024-03-1727.D. Secret Passwords2024-03-1928.F. Maximum White Subtree2024-03-1929.P3478 [POI2008] STA-Station2024-03-1930.P1347 排序2024-03-2131.P1960 郁闷的记者2024-03-2132.E1. Weights Division (easy version)2024-03-2233.P5007 DDOSvoid 的疑惑2024-03-2534.P2850 [USACO06DEC] Wormholes G2024-03-2635.P1265 公路修建2024-03-2836.P1354 房间最短路问题2024-03-2937.P2168 [NOI2015] 荷马史诗2024-03-3038.P8306 【模板】字典树2024-03-3039.P1481 魔族密码2024-03-3040.P3128 [USACO15DEC] Max Flow P2024-04-0241.P5536 【XR-3】核心城市2024-04-0242.P5836 [USACO19DEC] Milk Visits S2024-04-0243.P3384 【模板】重链剖分/树链剖分2024-04-0344.P5960 【模板】差分约束2024-04-0445.P7771 【模板】欧拉路径2024-04-0546.六度分离2024-04-0847.整数区间2024-04-0848.F. Alex's whims2024-04-1249.J. 上学2024-05-1450.Game on Tree2024-05-1451.E. We Need More Bosses2024-05-1552.B. Omkar and Heavenly Tree2024-05-1653.B. Mahmoud and Ehab and the bipartiteness2024-05-1654.P1668 [USACO04DEC] Cleaning Shifts S2024-05-1755.P6154 游走2024-05-2056.P8655 [蓝桥杯 2017 国 B] 发现环2024-05-2457.P10298 [CCC 2024 S4] Painting Roads2024-05-2458.P9650 [SNCPC2019] Escape Plan2024-05-2759.P9327 [CCC 2023 S4] Minimum Cost Roads2024-05-2960.P9026 [CCC2021 S4] Daily Commute2024-05-2961.P8724 [蓝桥杯 2020 省 AB3] 限高杆2024-06-0362.P4878 [USACO05DEC] Layout G2024-06-0463.P5663 [CSP-J2019] 加工零件2024-06-0464.P2731 [USACO3.3] 骑马修栅栏 Riding the Fences2024-06-0865.I. Disks2024-06-1766.P1351 [NOIP2014 提高组] 联合权值2024-06-2667.B. Time Travel2024-06-2968.F. Minimum Maximum Distance2024-07-1469.A. Book2024-07-1970.P1407 [国家集训队] 稳定婚姻2024-07-2071.P1991 无线通讯网2024-07-2372.P4047 [JSOI2010] 部落划分2024-07-23

73.P3275 [SCOI2011] 糖果2024-07-23

74.P1989 无向图三元环计数2024-07-2775.P1967 [NOIP2013 提高组] 货车运输2024-07-3176.D. Vitaly and Cycle2024-08-0177.P10838 『FLA - I』庭中有奇树2024-08-0778.P9751 [CSP-J 2023] 旅游巴士2024-08-1179.D. Colored Portals2024-08-21

收起

原题链接

题解

缩点+差分约束，求最小值故跑最长路

无解的情况：存在正权环，由于是有向图，所以环上的元素在一个强连通分量内，判断强连通分量内存不存在有正权值的边，然后缩点，拓扑跑一圈

缩点时要一个超级源点就可以只dfs一次了

code

#include <bits/stdc++.h>
using namespace std;

#define ll long long

const ll MAXN = 100005;

vector<pair<ll, ll>> G[MAXN], E[MAXN];
ll vis[MAXN], dfn[MAXN], low[MAXN], belong[MAXN], in[MAXN], val[MAXN];
stack<ll> st;
ll cnt, num;

void dfs(ll now) {
    vis[now] = 1;
    dfn[now] = low[now] = ++cnt;
    st.push(now);

    for (auto next : G[now]) {
        ll to = next.first;
        if (!dfn[to]) {
            dfs(to);
            low[now] = min(low[now], low[to]);
        } else if (!belong[to]) {
            low[now] = min(low[now], dfn[to]);
        }
    }

    if (low[now] == dfn[now]) {
        ll x;
        num++;
        do {
            x = st.top();
            st.pop();
            belong[x] = num;
        } while (x != now);
    }
}

void solve() {
    ll n, k;
    cin >> n >> k;

    for (ll i = 1; i <= k; i++) {
        ll op, x, y;
        cin >> op >> x >> y;

        if (op == 1) {
            G[x].push_back({y, 0});
            G[y].push_back({x, 0});
        } else if (op == 2) {
            G[x].push_back({y, 1});
        } else if (op == 3) {
            G[y].push_back({x, 0});
        } else if (op == 4) {
            G[y].push_back({x, 1});
        } else {
            G[x].push_back({y, 0});
        }
    }

    for (ll i = 1; i <= n; i++) G[0].push_back({i, 1});
    dfs(0);

    bool flag = true;
    for (ll i = 1; i <= n; i++) {
        for (auto next : G[i]) {
            ll x = belong[i], y = belong[next.first];
            if (x == y && next.second != 0) {
                flag = false;
            }
            if (x != y) {
                E[x].push_back({y, next.second});
                in[y]++;
            }
        }
    }

    if (!flag) {
        cout << -1 << endl;
        return;
    }

    queue<ll> q;
    for (ll i = 1; i <= num; i++) {
        val[i] = 1;
        if (!in[i]) q.push(i);
    }

    while (!q.empty()) {
        ll now = q.front();
        q.pop();

        for (auto next : E[now]) {
            ll to = next.first;
            val[to] = max(val[to], val[now] + next.second);
            in[to]--;
            if (!in[to]) {
                q.push(to);
            }
        }
    }

    ll ans = 0;
    for(ll i = 1; i <= n; i++) {
        ans += val[belong[i]];
    }
    cout << ans << endl;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    solve();
    return 0;
}