【网络流】 HDOJ 3395 Special Fish

这道题求的是最大费用流。。。不是最大费用最大流。。。还有求最大费用就是把求最小费用反过来就行了。。。

#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 205#define maxm 50005#define eps 1e-10#define mod 1000000007#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#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;}void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}// 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 vis[maxn];int cap[maxn];int dis[maxn];int cur[maxn];int flow, cost, s, t, T;char g[maxn][maxn];int val[maxn];int n;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(void){memset(dis, INF, sizeof dis);cur[s] = -1;vis[s] = ++T;dis[s] = 0;cap[s] = INF;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] == INF) return false;flow += cap[t];cost += cap[t] * dis[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 mcmf(void){flow = cost = 0;while(spfa());return -cost;}void init(void){cntE = T = 0;memset(H, -1, sizeof H);memset(vis, 0, sizeof vis);}void read(void){for(int i = 1; i <= n; i++) scanf("%d", &val[i]);for(int i = 1; i <= n; i++) scanf("%s", g[i] + 1);}void work(void){s = 0, t = 2 * n + 1;for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++) {if(i == j) continue;if(g[i][j] == '1')addedges(i, n + j, 1, -(val[i] ^ val[j]));}for(int i = 1; i <= n; i++) addedges(i, t, 1, 0);for(int i = 1; i <= n; i++) addedges(s, i, 1, 0);for(int i = 1; i <= n; i++) addedges(n + i, t, 1, 0);printf("%d\n", mcmf());}int main(void){while(scanf("%d", &n), n != 0) {init();read();work();}return 0;}