Drainage Ditches POJ - 1273 （网络流，最大流）

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.

Input

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.

Output

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

Sample Input

Sample Output

最裸的网络流算法，从1 到 n，最大流是多少，

Dinic 这个分层算法，

#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 500;
struct node{
    int next,v,w;
}edge[MAXN];
int head[MAXN],cur[MAXN],dep[MAXN];
int sign,n,m,s,t;
void add_edge(int u,int v, int w){
    sign++;
    edge[sign].next = head[u];
    head[u] = sign;
    edge[sign].v = v;
    edge[sign].w = w;
}
void init(){
    memset(head,-1,sizeof(head));
    s = 1; t = n;
    int x,y,z;
    sign = -1;
    for (int i = 0; i < m; i++){
        scanf("%d%d%d",&x,&y,&z);
        add_edge(x,y,z);
        add_edge(y,x,0);
    }
}

bool bfs(){
    queue<int>Q;
    while(!Q.empty()) Q.pop();
    Q.push(s);
    memset(dep,0,sizeof(dep));
    dep[s] = 1;
    while(!Q.empty()){
        int u = Q.front();
        Q.pop();
        for (int i = head[u]; i != -1; i = edge[i].next){
            int v = edge[i].v;
            if (dep[v]==0 && edge[i].w > 0)
            {
                dep[v] = dep[u]+1;
                if (v == t) return 1; //这是个优化，如果提前搜到 汇点，就退出来。
                Q.push(v);
            }
        }
    }
    if (dep[t]) return 1;
    return 0;
}

int dfs(int u,int flow){
    if (u == t || flow == 0) return flow; //我自己觉的，这个flow 可以优化，所以才 加上 flow == 0；
    for (int &i = cur[u]; i != -1; i = edge[i].next){  //&i  这个是引用，同时变。避免走重复的边。
        int v = edge[i].v,w = edge[i].w;
        if (dep[v] > dep[u] && w > 0){ 
            int d = dfs(v,min(flow,w));
            if (d){
                edge[i].w -= d; //正向边 减去流量，
                edge[i^1].w += d; //反向边 加上流量。
                return d;
            }
        }
    }
    return 0;
}

int Dinic(){
    int Ans=0,d;
    while(bfs()){
        for (int i = 1; i <= n; i++)
            cur[i] = head[i];  // cur 代表当前弧优化。一开始要初始化。
        while( d = dfs(s,INF)) {  //每次dfs返回一个最小的流量，然后接着搜dfs 直到dfs的结果为0，然后在bfs重新建深度。
            Ans += d;
        }
    }
    return Ans;
}

int main() {
    while(scanf("%d%d",&m,&n) == 2) {
        init();
        printf("%d\n", Dinic());
    }
    return 0;
}

Drainage Ditches POJ - 1273 （网络流，最大流）

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