CodeForces903G Yet Another Maxflow Problem 扫描线 + 线段树 + 最小割

给定两条链(A, B)，其中(A)链某些点向(B)链有连边，支持修改(A)链中的某条边权以及查询(A_1)到(B_n)的最大流

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显而易见，(A)和(B)链中一定满足左部分属于(S)集，右部分属于(T)集

枚举(A, B)的分界点在哪里，我们就能知道哪些边需要被割掉

可以发现，对于(A)链上的一个点而言，割(A,, B)之间的边以及(B)边的最小值是确定的

那么，对于(A)链上的每个点用扫描线预处理出这个最小值

然后最后再来一个线段树来维护(A)链上的权值即可

复杂度(O(n log n))

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#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

#define ll long long
#define ri register int
#define rep(io, st, ed) for(ri io = st; io <= ed; io ++)
#define drep(io, ed, st) for(ri io = ed; io >= st; io --)
    
#define gc getchar
inline int read() {
    int p = 0, w = 1; char c = gc();
    while(c > '9' || c < '0') { if(c == '-') w = -1; c = gc(); }
    while(c >= '0' && c <= '9') p = p * 10 + c - '0', c = gc();
    return p * w;
}

const int sid = 2e5 + 5;

int n, m, q;
int X[sid], Y[sid];

struct myk {
    int a, b, v;
    friend bool operator < (myk x, myk y)
    { return x.a < y.a; }
} Q[sid];

#define ls (o << 1)
#define rs (o << 1 | 1)

struct Kujuo_Miyako_Saiko {
    
    ll mi[sid << 2], add[sid << 2];
    
    inline void build(int o, int l, int r) {
        if(l == r) { mi[o] = Y[l - 1]; return; }
        int mid = (l + r) >> 1;
        build(ls, l, mid);
        build(rs, mid + 1, r);
        mi[o] = min(mi[ls], mi[rs]);
    }
    
    inline void mdf(int o, int l, int r, int ml, int mr, ll v) {
        if(ml > r || mr < l) return;
        if(ml <= l && mr >= r) { mi[o] += v; add[o] += v; return; }
        int mid = (l + r) >> 1;
        mdf(ls, l, mid, ml, mr, v);
        mdf(rs, mid + 1, r, ml, mr, v);
        mi[o] = min(mi[ls], mi[rs]) + add[o];
    }
    
} km;

ll V[sid];
inline void solve1() {
    km.build(1, 1, n);
    sort(Q + 1, Q + m + 1);
    for(ri i = 1, j = 1; i <= n; i ++) {
        while(Q[j].a == i && j <= m) km.mdf(1, 1, n, 1, Q[j].b, Q[j].v), j ++;
        V[i] = km.mi[1];
    }
}

ll mi[sid << 2];

inline void build(int o, int l, int r) {
    if(l == r) { mi[o] = V[l] + X[l]; return; }
    int mid = (l + r) >> 1;
    build(ls, l, mid); 
    build(rs, mid + 1, r);
    mi[o] = min(mi[ls], mi[rs]);
}

inline void mdf(int o, int l, int r, int p, int v) {
    if(l == r) { mi[o] = V[l] + v; return; }
    int mid = (l + r) >> 1;
    if(p <= mid) mdf(ls, l, mid, p, v);
    else mdf(rs, mid + 1, r, p, v);
    mi[o] = min(mi[ls], mi[rs]);
}

inline void solve2() {
    build(1, 1, n);
    printf("%lld
", mi[1]);
    rep(i, 1, q) {
        int v = read(), w = read();
        mdf(1, 1, n, v, w); printf("%lld
", mi[1]);
    }
}

int main() {
    n = read(); m = read(); q = read();
    rep(i, 1, n - 1) X[i] = read(), Y[i] = read();
    rep(i, 1, m) Q[i].a = read(), Q[i].b = read(), Q[i].v = read();
    solve1(); solve2();
    return 0;
}