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.
#include<bits/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,rt<<1 #define rson m+1,r,rt<<1|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 pair<int,int> pll; const int inf = 0x3f3f3f3f; const int _inf = 0xc0c0c0c0; const LL INF = 0x3f3f3f3f3f3f3f3f; const LL _INF = 0xc0c0c0c0c0c0c0c0; const LL mod = (int)1e9+7; 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){queue<int> 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)*m+j, (i-2)*m+j, a[2][i-1][j] + a[3][i-1][j], 1);if(j > 1) Flow.Eadd((i-1)*m+j, (i-1)*m+j-1, a[4][i][j-1] + a[5][i][j-1], 1);}}ans -= Flow.Dinic(s, t) / 2;printf("%d\n", ans);return 0; }
