BZOJ 2400 Spoj 839 Optimal Marks

最小割

显然按位来做。我们可以发现我们实际上是在把所有点划分成两个集合，两个集合之间可能有交叉的代价，这就类似于最小割模型！要求点权和最小可以把代价放大到10000并且将所有待定点与s（表示0集合）连一条代价1的边，这样就可以有主次关系了。

#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#define N 505#define M 2005#define ll long longusing namespace std;struct edge{int next,to,flow;}e[M*10];const int INF = 1<<29, mod = 10000;int ecnt = 1, last[N], cur[N], n, m, k , S, T, label[N], v[N];bool d[N][N];void addedge(int a, int b, int c){    e[++ecnt]=(edge){last[a],b,c};    last[a]=ecnt;    e[++ecnt]=(edge){last[b],a,0};    last[b]=ecnt;}void clr(){    S = 501;    T = 502;    memset(d,0,sizeof(d));    memset(v,0,sizeof(v));}void build(int pos){    ecnt=1;    memset(last,0,sizeof(last));    int pre = 1<<pos;    for(int i = 1; i <= n; i++)    {        if(v[i] == -1)addedge(S,i,1);        else if(v[i] & pre)addedge(i,T,INF);        else addedge(S,i,INF);        for(int j = 1; j <= n; j++)            if(d[i][j])                addedge(i,j,mod);    }}bool bfs(){    memset(label,0,sizeof(label));    queue<int> q;    q.push(S);    label[S]=1;    while(!q.empty())    {        int x = q.front();        q.pop();        for(int i = last[x]; i; i=e[i].next)        {            int y=e[i].to;            if(label[y] || !e[i].flow)continue;            label[y] = label[x] + 1;            if(y==T)return 1;            q.push(y);        }    }    return 0;}int dfs(int x, int flow){    if(x==T)return flow;    int use = 0;    for(int &i = cur[x]; i; i=e[i].next)    {        int y=e[i].to;        if(label[y] != label[x] + 1)continue;        int w = dfs(y, min(e[i].flow, flow-use));        e[i].flow -= w;        e[i^1].flow += w;        use += w;        if(use == flow)return use;    }    return use;}int dinic(){    int ret = 0;    while(bfs())    {        memcpy(cur,last,sizeof(cur));        ret += dfs(S,INF);    }    return ret;}int main(){    clr();    scanf("%d%d",&n,&m);    ll sum = 0;    for(int i = 1, x; i <= n; i++)    {        scanf("%d",&x);        if(x<0)v[i]=-1;        else v[i]=x, sum+=x;    }    for(int i = 1, a, b; i <= m; i++)    {        scanf("%d%d",&a,&b);        d[a][b]=d[b][a]=1;    }    ll eans = 0, pans = 0, ans;    for(int i = 0; i < 32; i++)    {        build(i);        ans = dinic();        eans += ans/mod * (1<<i);        pans += ans%mod * (1<<i);    }    pans += sum;    printf("%lld\n%lld\n",eans,pans);}