D - Nanami‘s Power Plant

思路：類似與bzoj切糕那道題的模型。。

#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 = 21000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-8;
const int MAX = 2e5;

int head[N], level[N], tot, S, T;
int n, m, a[N], b[N], c[N], L[N], R[N];
struct node {
    int to, w, nx;
} edge[
 
2000007];

void add(int u, int v, int w) {
    edge[tot].to = v;
    edge[tot].w = w;
    edge[tot].nx = head[u];
    head[u] = tot++;

    edge[tot].to = u;
    edge[tot].w = 0;
    edge[tot].nx = head[v];
    head[v] = tot++;
}


bool bfs() {
    memset(level, 0, sizeof(level));
    queue
 
<int> que; level[S] = 1; que.push(S);
    while(!que.empty()) {
        int u = que.front(); que.pop();
        if(u == T) return true;
        for(int i = head[u]; ~i; i = edge[i].nx) {
            int v = edge[i].to, w = edge[i].w;
            if(level[v] || w <= 0) continue;
            level[v] = level[u] + 1;
            que.push(v);
        }
    }
    return false;
}

int dfs(int u, int p) {
    if(u == T) return p;
    int ret = 0;
    for(int i = head[u]; ~i; i = edge[i].nx) {
        int v = edge[i].to, w = edge[i].w;
        if(level[v] != level[u] + 1 || w <= 0) continue;
        int f = dfs(v, min(w, p - ret));
        ret += f;
        edge[i].w -= f;
        edge[i ^ 1].w += f;
        if(ret == p) break;
    }
    if(!ret) level[u] = 0;
    return ret;
}

int Dinic() {
    int ans = 0;
    while(bfs()) ans += dfs(S, inf);
    return ans;
}
void init() {
    memset(head, -1, sizeof(head));
    tot = 0;
}

inline int ID(int x, int y) {
    return x*208+y+102;
}
inline int getVal(int who, int x) {
    return a[who]*x*x + b[who]*x + c[who];
}

int main() {
    init();
    S = 21000, T = 0;
    cin >> n >> m;
    for(int i = 1; i <= n; i++) cin >> a[i] >> b[i] >> c[i];
    for(int i = 1; i <= n; i++) {
        cin >> L[i] >> R[i];
        add(S, ID(i, L[i]-1), inf);
        for(int j = L[i]; j <= R[i]; j++)
            add(ID(i, j-1), ID(i, j), MAX-getVal(i, j));
        add(ID(i, R[i]), T, inf);
    }
    while(m--) {
        int u, v, d;
        cin >> u >> v >> d;
        for(int j = L[u]; j <= R[u]; j++)
            if(j-d>=L[v] && j-d-1<=R[v])
                add(ID(u, j-1), ID(v, j-d-1), inf);
    }
    printf("%d\n", n*MAX-Dinic());
    return 0;
}

/*
*/

Codeforces Round #248 (Div. 1) D - Nanami's Power Plant 最小割