【hdu3820】【最小割】Golden Eggs

很巧妙的建图方法。

一开始只想到将单个颜色的做法，对于多个颜色就没想出来怎么建图了。

大体思路还是二分图最大独立集的方法。

首先黑白染色，分成黑白两个集合。设金的价值为val1,银的价值为val2

由于有两种颜色，考虑将每个点拆成两个点k,k'

对于白色集合的点，k代表金，k'代表银；对于黑色集合的点，k代表银，k'代表金

白色集合的点:

s -> k，流量val1 ; k' -> t，流量val2

k -> k' ，流量为inf

黑色集合的点:

s -> k，流量为val2 ; k' -> t，流量val1

k -> k'，流量为inf

每个白色集合的点k，与每个黑色集合的点 k' 连边，容量为g

每个黑色集合的点k , 与每个白色集合的点 k' 连边， 容量为s

最终答案为总价值 - 最小割

代码:

#include<cstdio>#include<cstring>using namespace std;const int maxg = 50 + 10;const int maxn = 5000 + 10;const int maxm = 200000;const int inf = 0x3f3f3f3f;struct Edge{int pos,c;int next;}E[maxm];int map1[maxg][maxg],map2[maxg][maxg],head[maxn];int dis[maxn],pre[maxn],cur[maxn],gap[maxn];int T,n,m,g,h;int NE,s,t,nodenum;int cas = 0,sum = 0;void init(){freopen("hdu3820.in","r",stdin);freopen("hdu3820.out","w",stdout);}void add(int u,int v,int c){E[NE].pos = v;E[NE].c = c;E[NE].next = head[u];head[u] = NE++;E[NE].pos = u;E[NE].c = 0;E[NE].next = head[v];head[v] = NE++;}void build_map(){memset(E,0,sizeof(E));memset(head,-1,sizeof(head));NE = 0;s = 0,t = n * m * 2 + 1;nodenum = t + 1;int p = n * m;for(int i = 1;i <= n;i++){for(int j = 1;j <= m;j++){add((i - 1) * m + j,(i - 1) * m + j + p,inf);if((i + j) & 1){add(s,(i - 1) * m + j,map2[i][j]);add((i - 1) * m + j + p,t,map1[i][j]);if(i > 1)add((i - 1) * m + j,(i - 2) * m + j + p,h);if(i < n)add((i - 1) * m + j,i * m + j + p,h);if(j > 1)add((i - 1) * m + j,(i - 1) * m + j + p - 1,h);if(j < m)add((i - 1) * m + j,(i - 1) * m + j + p + 1,h);}else{add(s,(i - 1) * m + j,map1[i][j]);add((i - 1) * m + j + p,t,map2[i][j]);if(i > 1)add((i - 1) * m + j,(i - 2) * m + j + p,g);if(i < n)add((i - 1) * m + j,i * m + j + p,g);if(j > 1)add((i - 1) * m + j,(i - 1) * m + j + p - 1,g);if(j < m)add((i - 1) * m + j,(i - 1) * m + j + p + 1,g);}}}}void checkmin(int &a,int b){if(a == -1 || a > b)a = b;}int sap(){memset(gap,0,sizeof(gap));memset(dis,0,sizeof(dis));for(int i = s;i <= t;i++)cur[i] = head[i];int u = pre[s] = s,maxflow = 0,aug = -1;gap[s] = nodenum;while(dis[s] < nodenum){loop:   for(int &i = cur[u];i != -1;i = E[i].next){int v = E[i].pos;if(E[i].c && dis[u] == dis[v] + 1){checkmin(aug,E[i].c);pre[v] = u;u = v;if(v == t){maxflow += aug;for(u = pre[u];v != s;v = u,u = pre[u]){E[cur[u]].c -= aug;E[cur[u]^1].c += aug;}aug = -1;}goto loop;}}int mind = nodenum;for(int i = head[u];i != -1;i = E[i].next){int v = E[i].pos;if(E[i].c && (mind > dis[v])){cur[u] = i;mind = dis[v];}}if((--gap[dis[u]]) == 0)break;gap[dis[u] = mind + 1]++;u = pre[u];}return maxflow;}void solve(){build_map();printf("Case %d: %d\n",++cas,sum - sap());}void readdata(){scanf("%d",&T);while(T--){memset(map1,0,sizeof(map1));memset(map2,0,sizeof(map2));scanf("%d%d%d%d",&n,&m,&g,&h);sum = 0;for(int i = 1;i <= n;i++){for(int j = 1;j <= m;j++){scanf("%d",&map1[i][j]);sum += map1[i][j];}}for(int i = 1;i <= n;i++){for(int j = 1;j <= m;j++){scanf("%d",&map2[i][j]);sum += map2[i][j];}}solve();}}int main(){init();readdata();return 0;}