"bzoj1449/bzoj2895" [JSOI2009] Team yield/team budget fee flow

Title Description

Input

Output

An integer that represents the minimum value of the sum of all the teams in the league.

Sample input

3 3
1 0 2 1
1 1 10 1
0 1 3 3
0 S
2 3
3 1

Sample output

43

Exercises

Cost flow

Because there is a win a lose, more difficult to calculate. We can assume that they all lose and then arrange the winning situation.

Set fi for I also to play the number of games, then the initial yield of ∑ci*wi^2+di* (LI+FI) ^2.

S-> per game with a capacity of 1 and a cost of 0.

The two teams for each match, with a capacity of 1, cost 0.

Because the cost is changed to include the square, so we need to split the side to do.

The group I team to the T-Fi bar, the capacity is 1, the J-side indicates win J Field more than win j-1 field more income, so the cost should be ci* (wi+j) ^2+di* (wi+fi-j) ^2-ci* (li+j-1) ^2-di* (li+j+1) ^2.

For convenience, the FI is added directly to Li.

Then run the minimum cost maximum flow, plus the previous initial gain is the answer.

#include <cstdio> #include <cstring> #include <queue> #define N 10010#define M 3500000using namespace  Std;queue<int> Q;int W[n], l[n], c[n], d[n], x[n], y[n], F[n];int head[n], to[m], val[m], cost[m], next[m] , CNT = 1, S, T, Dis[n], from[n], pre[n];void Add (int x, int y, int v, int c) {to[++cnt] = y, val[cnt] = V, cost [CNT] = c, next[cnt] = head[x], head[x] = cnt;to[++cnt] = x, val[cnt] = 0, cost[cnt] =-C, next[cnt] = Head[y], head [Y] = cnt;} BOOL SPFA () {int x, I;memset (from,-1, sizeof (from)), memset (DIS, 0x3f, sizeof (DIS));d is[s] = 0, Q.push (s), and while (!q.em Pty ()) {x = Q.front (), Q.pop (); for (i = head[x]; i; i = Next[i]) if (Val[i] && dis[to[i]] > Dis[x] + cost[i]) dis [To[i]] = dis[x] + cost[i], from[to[i] [= x, pre[to[i]] = i, Q.push (To[i]);} return ~from[t];} int Mincost () {int i, k, ans = 0;while (SPFA ()) {k = 0x7fffffff;for (i = t; I! = s; i = from[i]) k = min (k, val[pre[i]); Ans + = k * Dis[t];for (i = t; I! = s; i = fRom[i]) val[pre[i] [= k, val[pre[i] ^ 1] + = k; return ans;} int main () {int n, m, I, j, ans = 0;scanf ("%d%d", &n, &m), s = 0, t = m + n + 1;for (i = 1; I <= n; i + +) scanf ("%d%d%d%d", &w[i], &l[i], &c[i], &d[i]); for (i = 1; I <= m; i + +) scanf ("%d%d", &x [i], &y[i]), F[x[i]] + +, F[y[i]] + +, L[x[i]] + +, L[y[i]] + +; for (i = 1; I <= n; i + +) ans + = c[i] * W[i] *  W[i] + d[i] * l[i] * L[I];FOR (i = 1; I <= m; i + +) Add (s, I, 1, 0), add (I, x[i] + m, 1, 0), add (I, y[i] + M  , 1, 0); for (i = 1; I <= n; i + +) for (j = 1; J <= F[i]; j + +) Add (i + M, T, 1, C[i] * (2 * w[i) + 2 * j-1) -D[i] * (2 * l[i]-2 * j + 1));p rintf ("%d\n", ans + mincost ()); return 0;}

"bzoj1449/bzoj2895" [JSOI2009] Team yield/team budget fee flow