SPOJ 839 Optimal Marks 最小割模型的转化（按位求最大流）

此题Amber的论文上还是有讲，建图的方法就不再赘述

题意描述：一个无向图，一些顶点权值已知而一些顶点权值未知，其中图中边的权值为其关联的两个顶点的异或值，现在让你在未知权值的顶点上填上权值后使得要求所有的边权之和最小，输出每个顶点的权值

关键是怎样输出方案。

我们按位进行网络流时，只需要到已知的最大标号的最大的一位即可。

然后对每一位做完最大流后，还是dfs残留网络，找不满流的边能到达的所有点即可。这能保证S集合中的个数最少。即保证了有多个方案下标号和最小的方案

#include <iostream>#include <algorithm>#include <cstring>#include <string>#include <cstdio>#include <cmath>#include <queue>#include <map>#include <set>#define eps 1e-5#define MAXN 555#define MAXM 55555#define INF 100000007using namespace std;typedef int type;struct node{    int v;    type c, f;    int next, r;}edge[MAXM];int dist[MAXN], nm[MAXN], src, des, n;int head[MAXN], e;void add(int x, int y, type c){    edge[e].v = y;    edge[e].c = c;    edge[e].f = 0;    edge[e].r = e + 1;    edge[e].next = head[x];    head[x] = e++;    edge[e].v = x;    edge[e].c = 0;    edge[e].f = 0;    edge[e].r = e - 1;    edge[e].next = head[y];    head[y] = e++;}void rev_BFS(){    int Q[MAXN], h = 0, t = 0;    for(int i = 1; i <= n; ++i)    {        dist[i] = MAXN;        nm[i] = 0;    }    Q[t++] = des;    dist[des] = 0;    nm[0] = 1;    while(h != t)    {        int v = Q[h++];        for(int i = head[v]; i != -1; i = edge[i].next)        {            if(edge[edge[i].r].c == 0 || dist[edge[i].v] < MAXN)continue;            dist[edge[i].v] = dist[v] + 1;            ++nm[dist[edge[i].v]];            Q[t++] = edge[i].v;        }    }}void init(){    e = 0;    memset(head, -1, sizeof(head));}type maxflow(){    rev_BFS();    int u;    type total = 0;    int cur[MAXN], rpath[MAXN];    for(int i = 1; i <= n; ++i)cur[i] = head[i];    u = src;    while(dist[src] < n)    {        if(u == des)     // find an augmenting path        {            type tf = INF;            for(int i = src; i != des; i = edge[cur[i]].v)                tf = min(tf, edge[cur[i]].c);            for(int i = src; i != des; i = edge[cur[i]].v)            {                edge[cur[i]].c -= tf;                edge[edge[cur[i]].r].c += tf;                edge[cur[i]].f += tf;                edge[edge[cur[i]].r].f -= tf;            }            total += tf;            u = src;        }        int i;        for(i = cur[u]; i != -1; i = edge[i].next)            if(edge[i].c > 0 && dist[u] == dist[edge[i].v] + 1)break;        if(i != -1)     // find an admissible arc, then Advance        {            cur[u] = i;            rpath[edge[i].v] = edge[i].r;            u = edge[i].v;        }        else        // no admissible arc, then relabel this vtex        {            if(0 == (--nm[dist[u]]))break;    // GAP cut, Important!            cur[u] = head[u];            int mindist = n;            for(int j = head[u]; j != -1; j = edge[j].next)                if(edge[j].c > 0)mindist = min(mindist, dist[edge[j].v]);            dist[u] = mindist + 1;            ++nm[dist[u]];            if(u != src)                u = edge[rpath[u]].v;    // Backtrack        }    }    return total;}int nt, m, xx[MAXM], yy[MAXM], num[MAXN], k;int ans[MAXN], vis[MAXN], base;void dfs(int u){    vis[u] = 1;    ans[u] += base;    for(int i = head[u]; i != -1; i = edge[i].next)        if(edge[i].c && !vis[edge[i].v])            dfs(edge[i].v);}int main(){    int T, u, w;    scanf("%d", &T);    while(T--)    {        scanf("%d%d", &nt, &m);        n = nt + 2;        src = nt + 1;        des = nt + 2;        for(int i = 1; i <= m; i++) scanf("%d%d", &xx[i], &yy[i]);        scanf("%d", &k);        int mx = 0;        memset(num, -1, sizeof(num));        for(int i = 1; i <= k; i++)        {            scanf("%d%d", &u, &w);            num[u] = w;            mx = max(mx, w);        }        memset(ans, 0, sizeof(ans));        base = 1;        while(mx)        {            init();            for(int i = 1; i <= m; i++) add(xx[i], yy[i], 1), add(yy[i], xx[i], 1);            for(int i = 1; i <= nt; i++)            {                if(num[i] == -1) continue;                if(num[i] & 1) add(src, i, INF);                else add(i, des, INF);                num[i] >>= 1;            }            maxflow();            memset(vis, 0, sizeof(vis));            dfs(src);            base *= 2;            mx >>= 1;        }        for(int i = 1; i <= nt; i++)            printf("%d\n", ans[i]);    }    return 0;}