BZOJ1520 [POI2006]Szk-Schools

裸的费用流啊。。。

建图：对于一个点p拆成两个p1和p2，S向p1连边，流量为1，费用为0；p2向T连边流量为1，费用为0

然后i1向a2到b2分别连边，不妨设i1向p2连边，流量为1，费用为|i - p| * ki

跑一下费用流，如果流量不为n，NIE！然后答案就是费用之和。。。

/**************************************************************
    Problem: 1520
    User: rausen
    Language: C++
    Result: Accepted
    Time:1296 ms
    Memory:2380 kb
****************************************************************/
 
#include <cstdio>
#include <algorithm>
 
using namespace std;
const int N = 405;
const int M = 1e5 + 5;
const int inf = 1e9;
 
struct edges {
    int next, to, f, cost;
    edges() {}
    edges(int _n, int _t, int _f, int _c) : next(_n), to(_t), f(_f), cost(_c) {}
} e[M];
  
int n, S, T;
int first[N], tot = 1;
int q[N], d[N], g[N];
bool v[N];
 
inline int read() {
    int x = 0, sgn = 1;
    char ch = getchar();
    while (ch < '0' || '9' < ch) {
        if (ch == '-') sgn = -1;
        ch = getchar();
    }
    while ('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return sgn * x;
}
 
inline void Add_Edges(int x, int y, int f, int c) {
    e[++tot] = edges(first[x], y, f, c), first[x] = tot;
    e[++tot] = edges(first[y], x, 0, -c), first[y] = tot;
}
  
inline int calc() {
    int flow = inf, x;
    for (x = g[T]; x; x = g[e[x ^ 1].to])
        flow = min(flow, e[x].f);
    for (x = g[T]; x; x = g[e[x ^ 1].to])
        e[x].f -= flow, e[x ^ 1].f += flow;
    return flow;
}
 
#define y e[x].to
bool spfa() {
    int x, now, l, r;
    for (x = 1; x <= T; ++x)
        d[x] = inf;
    d[S] = 0, v[S] = 1, q[0] = S;
    for(l = r = 0; l != (r + 1) % N; ) {
        now = q[l], ++l %= N;
        for (x = first[now]; x; x = e[x].next) {
            if (d[now] + e[x].cost < d[y] && e[x].f) {
                d[y] = d[now] + e[x].cost, g[y] = x;
                if (!v[y]) {
                    v[y] = 1;
                    if (d[y] < d[q[l]])
                        q[(l += N - 1) %= N] = y;
                    else q[++r %= N] = y;
                }
            }
        }
        v[now] = 0;
    }
    return d[T] != inf;
}
#undef y
 
inline int work() {
    static int res, tot;
    res = 0, tot = 0;
    while (spfa()) {
        tot += calc();
        res += d[T];
    }
    if (tot == n) printf("%d
", res);
    else puts("NIE");
}
 
int main() {
    int i, j, a, b, k, m;
    n = read(), S = 2 * n + 1, T = S + 1;
    for (i = 1; i <= n; ++i)
        Add_Edges(S, i, 1, 0), Add_Edges(i + n, T, 1, 0);
    for (i = 1; i <= n; ++i) {
        m = read(), a = read(), b = read(), k = read();
        for (j = a; j <= b; ++j)
            Add_Edges(i, n + j, 1, abs((j - m) * k));
    }
    work();
    return 0;
}