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 = ;
 const int M = 1e5 + ;
 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 = ;
 int q[N], d[N], g[N];
 bool v[N];

 inline int read() {
     int x = , sgn = ;
     char ch = getchar();
     while (ch < '' || '' < ch) {
         if (ch == '-') sgn = -;
         ch = getchar();
     }
     while ('' <= ch && ch <= '') {
         x = x *  + ch - '';
         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, , -c), first[y] = tot;
 }

 inline int calc() {
     int flow = inf, x;
     for (x = g[T]; x; x = g[e[x ^ ].to])
         flow = min(flow, e[x].f);
     for (x = g[T]; x; x = g[e[x ^ ].to])
         e[x].f -= flow, e[x ^ ].f += flow;
     return flow;
 }

 #define y e[x].to
 bool spfa() {
     int x, now, l, r;
     for (x = ; x <= T; ++x)
         d[x] = inf;
     d[S] = , v[S] = , q[] = S;
     for(l = r = ; l != (r + ) % 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] = ;
                     if (d[y] < d[q[l]])
                         q[(l += N - ) %= N] = y;
                     else q[++r %= N] = y;
                 }
             }
         }
         v[now] = ;
     }
     return d[T] != inf;
 }
 #undef y

 inline int work() {
     static int res, tot;
     res = , tot = ;
     while (spfa()) {
         tot += calc();
         res += d[T];
     }
     if (tot == n) printf("%d\n", res);
     else puts("NIE");
 }

 int main() {
     int i, j, a, b, k, m;
     n = read(), S =  * n + , T = S + ;
     for (i = ; i <= n; ++i)
         Add_Edges(S, i, , ), Add_Edges(i + n, T, , );
     for (i = ; i <= n; ++i) {
         m = read(), a = read(), b = read(), k = read();
         for (j = a; j <= b; ++j)
             Add_Edges(i, n + j, , abs((j - m) * k));
     }
     work();
     return ;
 }