Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie's clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.
Farmer John knows not only how many gallons of water each
ditch can transport per minute but also the exact layout of the ditches,
which feed out of the pond and into each other and stream in a
potentially complex network.

Given all this information, determine the maximum rate at
which water can be transported out of the pond and into the stream. For
any given ditch, water flows in only one direction, but there might be a
way that water can flow in a circle.

The input includes several cases. For each case,
the first line contains two space-separated integers, N (0 <= N <=
200) and M (2 <= M <= 200). N is the number of ditches that
Farmer John has dug. M is the number of intersections points for those
ditches. Intersection 1 is the pond. Intersection point M is the stream.
Each of the following N lines contains three integers, Si, Ei, and Ci.
Si and Ei (1 <= Si, Ei <= M) designate the intersections between
which this ditch flows. Water will flow through this ditch from Si to
Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water
will flow through the ditch.

For each case, output a single integer, the maximum rate at which water may emptied from the pond.

5 4

1 2 40

1 4 20

2 4 20

2 3 30

3 4 10

50

题意 ： 给你一个源点和一个汇点，再给你一些中间边，同时给你他们边上的容量，求从源点到汇点最大流量是多少？
思路分析 ：网络流板子题
代码示例 ：

using namespace std;

#define ll long long

const int maxn = 205;

const int mod = 1e9+7;

const double eps = 1e-9;

const double pi = acos(-1.0);

const int inf = 0x3f3f3f3f;

int n, m;

struct node

{

    int next, v, flow; // flow可以理解为容量限制

}e[maxn<<1];

int cnt;

int head[maxn];

int aim; // 目标点

int deep[maxn]; // 分层图的深度

void addedge(int u, int v, int cap){

    e[cnt].v = v;

    e[cnt].flow = cap;

    e[cnt].next = head[u];

    head[u] = cnt++;

}

int que[10000];

bool bfs(int s, int t){

    memset(deep, 0, sizeof(deep));

    deep[s] = 1; que[0] = s; int head1 = 0, tail1 = 1;

    while(head1 < tail1){

        int u = que[head1++];

        for(int i = head[u]; i != -1; i = e[i].next){

            int v = e[i].v;

            if (!deep[v] && e[i].flow){ // 判断当前的边如果还可以流

                deep[v] = deep[u]+1;

                que[tail1++] = v;

            }

        }

    }

    return deep[t];

}

int dfs(int u, int f1){

    if (u == aim || f1 == 0) return f1;  // 这个优化非常的棒

    int f = 0;

    for(int i = head[u]; i != -1; i = e[i].next){ // 多路增广，利用dfs的特性

        int v = e[i].v;

        if (e[i].flow && deep[v] == deep[u]+1){

            int x = dfs(e[i].v, min(f1, e[i].flow));

            e[i].flow -= x; e[i^1].flow += x;

            f1 -= x; f += x;
　　　　　　  if (f1 == 0) return f; // !!!

        }

    }

    if (!f) deep[u] = -2; // 炸点优化,若当前点的流量为 0,则在此次中没有必要再去访问该点了

    return f;

}

int maxflow(int s, int t){

    aim = t; int ret = 0;

    cnt = 0;

    while(bfs(s, t)){

        ret += dfs(s, inf);

    }

    return ret;

}

int main() {

    //freopen("in.txt", "r", stdin);

    //freopen("out.txt", "w", stdout);

    int u, v, w;

    while(~scanf("%d%d", &n, &m)){

        cnt = 0;

        memset(head, -1, sizeof(head));

        for(int i = 1; i <= n; i++){

            scanf("%d%d%d", &u, &v, &w);

            addedge(u, v, w);

            addedge(v, u, 0); // 反向边的建立，并赋值 0

        }

        printf("%d\n", maxflow(1, m));

    }

    return 0;

}