POj 1273 Drainage Ditches （最大流 Dinic + ISAP 模板）

Drainage Ditches

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 76183 Accepted: 29594

Description

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

5 41 2 401 4 202 4 202 3 303 4 10

Sample Output

50

Source

USACO 93

题意： n个池塘， m个管道， 每个管道有一个容量值， 1位源点， n为汇点，问你n的最大流是多少

纯模板题~

第一道网络流~

一如网络流深似海的生涯开启...

Dinic板子

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>#include <vector>#include <queue>using namespace std;const int maxn = 205;const int INF = 2e9;int head[maxn], k, n, m, s, t, cur[maxn], d[maxn];struct node{    int v, w, next;}edge[maxn*2];void addEdge(int u, int v, int w){    edge[k].v = v;    edge[k].w = w;    edge[k].next = head[u];    head[u] = k++;}int bfs(){    memset(d, 0, sizeof(d));    d[s] = 1;    queue<int> q;    q.push(s);    while(!q.empty())    {        int u = q.front();        if(u == t) return 1;        q.pop();        for(int i = head[u]; i != -1; i = edge[i].next)        {            int to = edge[i].v, w = edge[i].w;            if(w && d[to] == 0)            {                d[to] = d[u] + 1;                q.push(to);            }        }    }    return 0;}int dfs(int u, int maxflow, int t){    if(t == u) return maxflow;    int ret = 0;    for(int &i = cur[u]; i != -1; i = edge[i].next)    {        int to = edge[i].v, w = edge[i].w;        if(w && d[to]==d[u]+1)        {            int f = dfs(to, min(maxflow-ret, w), t);            edge[i].w -= f;            edge[i^1].w += f;            ret += f;            if(ret == maxflow) return ret;        }    }    return ret;}int Dinic(){    int ans = 0;    while(bfs() == 1)    {        memcpy(cur, head, sizeof(head));        ans += dfs(s, INF, t);    }    return ans;}int main(){    while(~scanf("%d%d", &n, &m))    {        int a, b, flow;        k = 1; s = 1; t = m;        memset(head, -1, sizeof(head));//        memset(head, -1, sizeof(head));        for(int i = 1; i <= n; i++)        {            scanf("%d%d%d", &a, &b, &flow);            addEdge(a, b, flow);            addEdge(b, a, 0);        }        printf("%d\n", Dinic());    }    return 0;}

ISAP板子

#include <stdio.h>    #include <string.h>    #include <algorithm>    #define clear(A, X) memset (A, X, sizeof A)    #define copy(A, B) memcpy (A, B, sizeof A)    using namespace std;        const int maxE = 1000000;    const int maxN = 100000;    const int maxQ = 1000000;    const int oo = 0x3f3f3f3f;         struct Edge {       int v;//弧尾       int c;//容量       int n;//指向下一条从同一个弧头出发的弧    } edge[maxE];//边组           int adj[maxN], cntE;//前向星的表头    int Q[maxQ], head, tail;//队列    int d[maxN], cur[maxN], pre[maxN], num[maxN];    int sourse, sink, nv;//sourse：源点，sink：汇点，nv：编号修改的上限    int n, m;        void addedge (int u, int v, int c) {//添加边        //正向边        edge[cntE].v = v;        edge[cntE].c = c;//正向弧的容量为c        edge[cntE].n = adj[u];        adj[u] = cntE++;                //反向边        edge[cntE].v = u;        edge[cntE].c = 0;//反向弧的容量为0        edge[cntE].n = adj[v];        adj[v] = cntE++;    }         void rev_bfs () {//反向BFS标号        clear (num, 0);        clear (d, -1);//没标过号则为-1        d[sink] = 0;//汇点默认为标过号        num[0] = 1;        head = tail = 0;        Q[tail++] = sink;             while (head != tail) {            int u = Q[head++];            for (int i = adj[u]; ~i; i = edge[i].n) {                int v = edge[i].v;                if (~d[v]) continue;//已经标过号                d[v] = d[u] + 1;//标号                Q[tail++] = v;                num[d[v]]++;            }        }    }        int ISAP() {        copy (cur, adj);//复制，当前弧优化        rev_bfs ();//只用标号一次就够了，重标号在ISAP主函数中进行就行了        int flow = 0, u = pre[sourse] = sourse, i;            while (d[sink] < nv) {//最长也就是一条链，其中最大的标号只会是nv - 1，如果大于等于nv了说明中间已经断层了。            if (u == sink) {//如果已经找到了一条增广路，则沿着增广路修改流量                int f = oo, neck;                for (i = sourse; i != sink; i = edge[cur[i]].v) {                    if (f > edge[cur[i]].c){                        f = edge[cur[i]].c;//不断更新需要减少的流量                        neck = i;//记录回退点，目的是为了不用再回到起点重新找                    }                }                for (i = sourse; i != sink; i = edge[cur[i]].v) {//修改流量                    edge[cur[i]].c -= f;                    edge[cur[i] ^ 1].c += f;                }                flow += f;//更新                u = neck;//回退            }            for (i = cur[u]; ~i; i = edge[i].n) if (d[edge[i].v] + 1 == d[u] && edge[i].c) break;            if (~i) {//如果存在可行增广路，更新                cur[u] = i;//修改当前弧                pre[edge[i].v] = u;                u = edge[i].v;            }            else {//否则回退，重新找增广路                if (0 == (--num[d[u]])) break;//GAP间隙优化，如果出现断层，可以知道一定不会再有增广路了                int mind = nv;                for (i = adj[u]; ~i; i = edge[i].n) {                    if (edge[i].c && mind > d[edge[i].v]) {//寻找可以增广的最小标号                        cur[u] = i;//修改当前弧                        mind = d[edge[i].v];                    }                }                d[u] = mind + 1;                num[d[u]]++;                u = pre[u];//回退            }        }                return flow;    }        void init () {//初始化        clear (adj, -1);        cntE = 0;    }        void work () {        int u, v, c;        init ();        for (int i = 0; i < m; ++ i) scanf ("%d%d%d", &u, &v, &c), addedge (u, v, c);        sourse = 1; sink = n; nv = sink + 1;        printf ("%d\n", ISAP ());    }      int main() {        while (~scanf("%d%d", &m, &n)) work ();        return 0;    }