HDU 3987 最小割模型

读完题后，就知道是最小割了，最小割=最大流，但是题目又说要最少边的最小割，输出边的个数

这样建边得时候就要换种方式了，将边权w变为w *(E + 1) + 1，这时候求出的最大流实际上附带了一个信息，就是边的个数，最后的结果直接mod （E+1 ）即可

/*ID: CUGB-wwjPROG:LANG: C++*/#include <iostream>#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <cctype>#include <string>#include <cstring>#include <cmath>#include <ctime>#define INF 11111111111111LL#define MAXN 1005#define MAXM 444444#define PI acos(-1.0)using namespace std;struct node{    int ver;    // vertex    long long cap;    // capacity    long long flow;   // current flow in this arc    int next, rev;}edge[MAXM];long long dist[MAXN];int numbs[MAXN], src, des, n, m;int head[MAXN], e;void add(int x, int y, long long  c){       //e记录边的总数    edge[e].ver = y;    edge[e].cap = c;    edge[e].flow = 0;    edge[e].rev = e + 1;        //反向边在edge中的下标位置    edge[e].next = head[x];   //记录以x为起点的上一条边在edge中的下标位置    head[x] = e++;           //以x为起点的边的位置    //反向边    edge[e].ver = x;    edge[e].cap = 0;  //反向边的初始网络流为0    edge[e].flow = 0;    edge[e].rev = e - 1;    edge[e].next = head[y];    head[y] = e++;}void rev_BFS(){    int Q[MAXN], qhead = 0, qtail = 0;    for(int i = 1; i <= n; ++i)    {        dist[i] = MAXN;        numbs[i] = 0;    }    Q[qtail++] = des;    dist[des] = 0;    numbs[0] = 1;    while(qhead != qtail)    {        int v = Q[qhead++];        for(int i = head[v]; i != -1; i = edge[i].next)        {            if(edge[edge[i].rev].cap == 0 || dist[edge[i].ver] < MAXN)continue;            dist[edge[i].ver] = dist[v] + 1;            ++numbs[dist[edge[i].ver]];            Q[qtail++] = edge[i].ver;        }    }}long long  maxflow(){    int u;    long long totalflow = 0;    int Curhead[MAXN], revpath[MAXN];    for(int i = 1; i <= n; ++i)Curhead[i] = head[i];    u = src;    while(dist[src] < n)    {        if(u == des)     // find an augmenting path        {            long long  augflow = INF;            for(int i = src; i != des; i = edge[Curhead[i]].ver)                augflow = min(augflow, edge[Curhead[i]].cap);            for(int i = src; i != des; i = edge[Curhead[i]].ver)            {                edge[Curhead[i]].cap -= augflow;                edge[edge[Curhead[i]].rev].cap += augflow;                edge[Curhead[i]].flow += augflow;                edge[edge[Curhead[i]].rev].flow -= augflow;            }            totalflow += augflow;            u = src;        }        int i;        for(i = Curhead[u]; i != -1; i = edge[i].next)            if(edge[i].cap > 0 && dist[u] == dist[edge[i].ver] + 1)break;        if(i != -1)     // find an admissible arc, then Advance        {            Curhead[u] = i;            revpath[edge[i].ver] = edge[i].rev;            u = edge[i].ver;        }        else        // no admissible arc, then relabel this vertex        {            if(0 == (--numbs[dist[u]]))break;    // GAP cut, Important!            Curhead[u] = head[u];            long long  mindist = n;            for(int j = head[u]; j != -1; j = edge[j].next)                if(edge[j].cap > 0)mindist = min(mindist, dist[edge[j].ver]);            dist[u] = mindist + 1;            ++numbs[dist[u]];            if(u != src)                u = edge[revpath[u]].ver;    // Backtrack        }    }    return totalflow;}void init(){    e = 0;    memset(head, -1, sizeof(head));}int main(){    int T;    scanf("%d", &T);    int x, y, fg;    long long z;    int cas = 0;    while(T--)    {        init();        scanf("%d%d", &n, &m);        for(int i = 0; i < m; i++)        {            scanf("%d%d%I64d%d", &x, &y, &z, &fg);            add(x + 1, y + 1, z * 222222LL + 1);            if(fg) add(y + 1, x + 1, z * 222222LL + 1);        }        src = 1;        des = n;        rev_BFS();        long long  ans = maxflow() % 222222LL;        printf("Case %d: %I64d\n", ++cas, ans);    }    return 0;}