CF434D Nanami's Power Plant 最小割

传送门

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因为连距离限制的边的细节调了贼久QAQ

这个题和HNOI2013 切糕性质相同，都是有距离限制的最小割问题

对于每一个函数，用一条链记录变量\(x\)在不同取值下这个函数的贡献。对于一个\(x_u \leq x_v + d\)的限制，用一条\(INF\)的边连接链上对应的两个点来限制这个条件。

注意一些细节的地方：

①注意每一个限制中每一个点对应另一条链的哪一个点，一定要想清楚；

②如果\(c<0\)，链上还会存在一些点不可能被割掉，还要连\(S\)和\(T\)相关的边限制这样的割。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<ctime>
#include<cctype>
#include<algorithm>
#include<cstring>
#include<iomanip>
#include<queue>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<vector>
#include<cmath>
#include<random>
#include<cassert>
#define int long long
#define inf 1e10
#define INF 1e13
//This code is written by Itst
using namespace std;

const int MAXN = 1e5 + 7 , MAXM = 1e6 + 7;
struct Edge{
    int end , upEd , f , c;
}Ed[MAXM];
int head[MAXN] , a[51] , b[51] , c[51] , l[51] , r[51] , ind[51][207];
int N , M , S , T , cntEd = 1;
queue < int > q;

inline void addEd(int a , int b , int c , int d = 0){
    Ed[++cntEd].end = b;
    Ed[cntEd].upEd = head[a];
    Ed[cntEd].f = c;
    Ed[cntEd].c = d;
    head[a] = cntEd;
}

inline void addE(int a , int b , int c , int d = 0 , bool f = 0){
    addEd(a , b , c , d); addEd(b , a , c * f , -d);
}

int cur[MAXN] , dep[MAXN];

inline bool bfs(){
    while(!q.empty())
        q.pop();
    q.push(S);
    memset(dep , 0 , sizeof(dep));
    dep[S] = 1;
    while(!q.empty()){
        int t = q.front();
        q.pop();
        for(int i = head[t] ; i ; i = Ed[i].upEd)
            if(Ed[i].f && !dep[Ed[i].end]){
                dep[Ed[i].end] = dep[t] + 1;
                if(Ed[i].end == T){
                    memcpy(cur , head , sizeof(head));
                    return 1;
                }
                q.push(Ed[i].end);
            }
    }
    return 0;
}

inline int dfs(int x , int mF){
    if(x == T)
        return mF;
    int sum = 0;
    for(int &i = cur[x] ; i ; i = Ed[i].upEd)
        if(Ed[i].f && dep[Ed[i].end] == dep[x] + 1){
            int t = dfs(Ed[i].end , min(mF - sum , Ed[i].f));
            if(t){
                Ed[i].f -= t;
                Ed[i ^ 1].f += t;
                sum += t;
                if(sum == mF)
                    break;
            }
        }
    return sum;
}

int Dinic(){
    int ans = 0;
    while(bfs())
        ans += dfs(S , INF);
    return ans;
}

inline int calc(int tp , int x){
    return a[tp] * x * x + b[tp] * x + c[tp];
}

signed main(){
#ifndef ONLINE_JUDGE
    freopen("in" , "r" , stdin);
    //freopen("out" , "w" , stdout);
#endif
    ios::sync_with_stdio(0);
    cin >> N >> M;
    for(int i = 1 ; i <= N ; ++i) cin >> a[i] >> b[i] >> c[i];
    T = 1e5;
    int cnt = 0;
    for(int i = 1 ; i <= N ; ++i){
        cin >> l[i] >> r[i];
        for(int j = 0 ; j <= 200 ; ++j) ind[i][j] = ++cnt;
        addE(S , ind[i][0] , l[i] == -100 ? inf - calc(i , -100) : INF);
        addE(ind[i][200] , T , INF);
        for(int j = 1 ; j <= 200 ; ++j)
            addE(ind[i][j - 1] , ind[i][j] , j - 100 >= l[i] && j - 100 <= r[i] ? inf - calc(i , j - 100) : INF);
    }
    while(M--){
        int a , b , c;
        cin >> a >> b >> c;
        for(int i = max(0ll , c) ; i <= min(200ll , 200 + c) ; ++i)
            addE(ind[a][i] , ind[b][i - c] , INF);
        if(c < 0){
            addE(S , ind[b][-c - 1] , INF);
            addE(ind[a][200 + c] , T , INF);
        }
    }
    cout << (int)(N * inf - Dinic());
    return 0;
}