现在百度怎么优化排名,seo的网站,大型网站开发 广州,项目投资网happiness 题解: 将图转换成最小割. 将割完的图中与S相连的点看做选文科, 与T相连的点看做选理科. flow(s, u) 文科值 flow(u,t) 理科值 假设u 和 v 一起选文科有奖励值z, flow(s,u) z/2 flow(s,v) z/2, flow(u,v) z/2 假设u 和 v 一起选理科有奖励值z, flow(u,t) z/2… happiness 题解: 将图转换成最小割. 将割完的图中与S相连的点看做选文科, 与T相连的点看做选理科. flow(s, u) 文科值 flow(u,t) 理科值 假设u 和 v 一起选文科有奖励值z,  flow(s,u) z/2  flow(s,v) z/2, flow(u,v) z/2 假设u 和 v 一起选理科有奖励值z,  flow(u,t) z/2  flow(v,t) z/2, flow(u,v) z/2 然后合并边. 具体理解可以画图看最小割之后的模型. 剩下的图才是价值. #includebits/stdc.h using namespace std; #define Fopen freopen(_in.txt,r,stdin); freopen(_out.txt,w,stdout); #define LL long long #define ULL unsigned LL #define fi first #define se second #define pb push_back #define lson l,m,rt1 #define rson m1,r,rt1|1 #define lch(x) tr[x].son[0] #define rch(x) tr[x].son[1] #define max3(a,b,c) max(a,max(b,c)) #define min3(a,b,c) min(a,min(b,c)) typedef pairint,int pll; const int inf 0x3f3f3f3f; const int _inf 0xc0c0c0c0; const LL INF 0x3f3f3f3f3f3f3f3f; const LL _INF 0xc0c0c0c0c0c0c0c0; const LL mod (int)1e97; int a[6][105][105]; int n, m; struct FFFlow{const static int N 105*105;static const int M N 3;int head[N], deep[N], cur[N];int w[M], to[M], nx[M];int tot;void add(int u, int v, int val){w[tot] val; to[tot] v;nx[tot] head[u]; head[u] tot;// w[tot] 0; to[tot] u; // nx[tot] head[v]; head[v] tot;}void Eadd(int u, int v, int val, int p){add(u, v, val);add(v, u, val * p);}int bfs(int s, int t){queueint q;memset(deep, 0, sizeof(deep));q.push(s);deep[s] 1;while(!q.empty()){int u q.front();q.pop();for(int i head[u]; ~i; i nx[i]){if(w[i] 0 deep[to[i]] 0){deep[to[i]] deep[u] 1;q.push(to[i]);}}}return deep[t] 0;}int Dfs(int u, int t, int flow){if(u t) return flow;for(int i cur[u]; ~i; i nx[i]){if(deep[u]1 deep[to[i]] w[i] 0){int di Dfs(to[i], t, min(w[i], flow));if(di 0){w[i] - di, w[i^1] di;return di;}}}return 0;}int Dinic(int s, int t){int ans 0, tmp;while(bfs(s, t)){for(int i 0; i t; i) cur[i] head[i];while(tmp Dfs(s, t, inf)) ans tmp;}return ans;}void init(){memset(head, -1, sizeof(head));tot 0;}}Flow; int main(){Flow.init();scanf(%d%d, n, m);int ans 0;for(int k 0; k 6; k){int x n, y m;if(k 2 || k 3) --x;if(k 5 || k 4) --y;for(int i 1; i x; i)for(int j 1; j y; j){scanf(%d, a[k][i][j]);ans a[k][i][j];}}int s 0, t n * m 1;for(int i 1; i n; i){for(int j 1; j m; j){Flow.Eadd(s, (i-1)*m j, a[0][i][j] * 2 a[2][i-1][j] a[2][i][j] a[4][i][j-1] a[4][i][j], 0);Flow.Eadd((i-1)*m j, t, a[1][i][j] * 2 a[3][i-1][j] a[3][i][j] a[5][i][j-1] a[5][i][j], 0);if(i 1) Flow.Eadd((i-1)*mj, (i-2)*mj, a[2][i-1][j] a[3][i-1][j], 1);if(j 1) Flow.Eadd((i-1)*mj, (i-1)*mj-1, a[4][i][j-1] a[5][i][j-1], 1);}}ans - Flow.Dinic(s, t) / 2;printf(%d\n, ans);return 0; } View Code 　　 转载于:https://www.cnblogs.com/MingSD/p/11254013.html