【2017 ACM/ICPC Asia Regional Qingdao Online 1009】hdu 6214 Smallest Minimum Cut

Smallest Minimum Cut

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 269    Accepted Submission(s): 76

Problem Description

Consider a network G=(V,E ) with source s and sink t. An s-t cut is a partition of nodes set V into two parts such that s and t belong to different parts. The cut set is the subset of E with all edges connecting nodes in different parts. A minimum cut is the one whose cut set has the minimum summation of capacities. The size of a cut is the number of edges in the cut set. Please calculate the smallest size of all minimum cuts.

 

Input

The input contains several test cases and the first line is the total number of casesT (1≤T≤300).
Each case describes a network G, and the first line contains two integers n (2≤n≤200) and m (0≤m≤1000) indicating the sizes of nodes and edges. All nodes in the network are labelled from1 to n.
The second line contains two different integers s and t (1≤s,t≤n) corresponding to the source and sink.
Each of the next m lines contains three integers u,v and w (1≤w≤255)  describing a directed edge from node u to v with capacity w.

 

Output

For each test case, output the smallest size of all minimum cuts in a line.

 

Sample Input

24 51 41 2 31 3 12 3 12 4 13 4 24 51 41 2 31 3 12 3 12 4 13 4 3

 

Sample Output

23

 

题意：

求最小割边数，模板题

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;const __int64 inf = 10000000000000LL;const int maxn = 100010;int head[maxn], tol, dep[maxn], n;struct node{    int from, to, next;    __int64 cap;} edge[1000000];void add(int u, int v, __int64 cap){    edge[tol] = (node)    {        u, v, head[u], cap    };    head[u] = tol++;    edge[tol] = (node)    {        v, u, head[v], 0    };    head[v] = tol++;}int bfs(int s, int t){    int que[maxn], front = 0, rear = 0;    memset(dep, -1, sizeof(dep));    dep[s] = 0;    que[rear++] = s;    while (front != rear)    {        int u = que[front++];        front %= maxn;        for (int i = head[u]; i != -1; i = edge[i].next)        {            int v = edge[i].to;            if (edge[i].cap > 0 && dep[v] == -1)            {                dep[v] = dep[u] + 1;                que[rear++] = v;                rear %= maxn;                if (v == t)                {                    return 1;                }            }        }    }    return 0;}__int64 dinic(int s, int t){    __int64 res = 0;    while (bfs(s, t))    {        int Stack[maxn], top, cur[maxn];        memcpy(cur, head, sizeof(head));        top = 0;        int u = s;        while (1)        {            if (t == u)            {                __int64 min = inf;                int loc;                for (int i = 0; i < top; i++)                    if (min > edge[Stack[i]].cap)                    {                        min = edge[Stack[i]].cap;                        loc = i;                    }                for (int i = 0; i < top; i++)                {                    edge[Stack[i]].cap -= min;                    edge[Stack[i] ^ 1].cap += min;                }                res += min;                top = loc;                u = edge[Stack[top]].from;            }            for (int i = cur[u]; i != -1; cur[u] = i = edge[i].next)                if (dep[edge[i].to] == dep[u] + 1 && edge[i].cap > 0)                {                    break;                }            if (cur[u] != -1)            {                Stack[top++] = cur[u];                u = edge[cur[u]].to;            }            else            {                if (top == 0)                {                    break;                }                dep[u] = -1;                u = edge[Stack[--top]].from;            }        }    }    return res;}int main(){    int i, j, m, T, s, t;    __int64 k;    scanf("%d", &T);    while (T--)    {        scanf("%d%d", &n, &m);        scanf("%d%d", &s, &t);        memset(head, -1, sizeof(head));        tol = 0;        while (m--)        {            scanf("%d%d%I64d", &i, &j, &k);            add(i, j, k * 200001 + 1);            //add(j, i, k * 200001 + 1);            add(j, i, 0);        }        __int64 ans = dinic(s, t);        cout << ans % 200001 << endl;    }    return 0;}