【网络流 FOJ 2143 Board Game

费用流。。。。拆边，把一条边拆成k条边，这样k条边的费用是单调递增的。然后奇偶建图。跑费用流的时候费用为正就退出。。。。

#include <iostream>#include <queue>#include <stack>#include <map>#include <set>#include <bitset>#include <cstdio>#include <algorithm>#include <cstring>#include <climits>#include <cstdlib>#include <cmath>#include <time.h>#define maxn 105#define maxm 100005#define eps 1e-7#define mod 1000000009#define INF 0x3f3f3f3f#define PI (acos(-1.0))#define lowbit(x) (x&(-x))#define mp make_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid #define rson o<<1 | 1, mid+1, R#define pii pair<int, int>#pragma comment(linker, "/STACK:16777216")typedef long long LL;typedef unsigned long long ULL;//typedef int LL;using namespace std;LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}//headstruct Edge{int v, c, w, next;Edge() {}Edge(int v, int c, int w, int next) : v(v), c(c), w(w), next(next) {}}E[maxm];queue<int> q;int H[maxn], cntE;int cap[maxn];int vis[maxn];int dis[maxn];int cur[maxn];int dir[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};int flow, cost, T, s, t;int n, m;void addedges(int u, int v, int c, int w){E[cntE] = Edge(v, c, w, H[u]);H[u] = cntE++;E[cntE] = Edge(u, 0, -w, H[v]);H[v] = cntE++;}bool spfa(){memset(dis, INF, sizeof dis);cur[s] = -1;vis[s] = ++T;cap[s] = INF;dis[s] = 0;q.push(s);while(!q.empty()) {int u = q.front();q.pop();vis[u] = T - 1;for(int e = H[u]; ~e; e = E[e].next) {int v = E[e].v, c = E[e].c, w = E[e].w;if(c && dis[v] > dis[u] + w) {dis[v] = dis[u] + w;cap[v] = min(cap[u], c);cur[v] = e;if(vis[v] != T) {vis[v] = T;q.push(v);}}}}if(dis[t] >= 0) return false;cost += cap[t] * dis[t];flow += cap[t];for(int e = cur[t]; ~e; e = cur[E[e ^ 1].v]) {E[e].c -= cap[t];E[e ^ 1].c += cap[t];}return true;}int mfmc(){flow = cost = 0;while(spfa());return cost;}void init(){cntE = T = 0;memset(H, -1, sizeof H);memset(vis, 0, sizeof vis);}inline int calc(int i, int j){return (i - 1) * m + j;}void work(int _){int kk, x;scanf("%d%d%d", &n, &m, &kk);int res = 0;s = 0, t = n * m + 1;for(int i = 1; i <= n; i++)for(int j = 1; j <= m; j++) {scanf("%d", &x);int tt = x * x;res += tt;if((i+j) % 2) {for(int k = 0; k < 4; k++) {int ti = i + dir[k][0];int tj = j + dir[k][1];if(ti <= 0 || ti > n || tj <= 0 || tj > m) continue;addedges(calc(i, j), calc(ti, tj), INF, 0);}}for(int k = 1; k <= kk; k++) {if((i+j) % 2) addedges(s, calc(i, j), 1, (x - k) * (x - k) - tt);else addedges(calc(i, j), t, 1, (x - k) * (x - k) - tt);tt = (x - k) * (x - k);}}printf("Case %d: %d\n", _, mfmc() + res);}int main(){int _;scanf("%d", &_);for(int i = 1; i <= _; i++) {init();work(i);}return 0;}