POJ1273-Drainage Ditches

Drainage Ditches

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 77789 Accepted: 30315

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：有M个池塘和N条水沟用来排水，池塘编号为1~M，1号池塘是所有水沟的源点，M号池塘是水沟的汇点。给你N条水沟所连接的池塘和所能流过的水量，求整个水沟从源点到汇点最多能流多少水

解题思路：很明显的一道网络流裸题

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;#define MAXN 500struct node{    int u, v, next, cap;} edge[MAXN*MAXN];int nt[MAXN], s[MAXN], d[MAXN], visit[MAXN];int cnt;void init(){    cnt = 0;    memset(s, -1, sizeof(s));}void add(int u, int v, int c){    edge[cnt].u = u;    edge[cnt].v = v;    edge[cnt].cap = c;    edge[cnt].next = s[u];    s[u] = cnt++;    edge[cnt].u = v;    edge[cnt].v = u;    edge[cnt].cap = 0;    edge[cnt].next = s[v];    s[v] = cnt++;}bool BFS(int ss, int ee){    memset(d, 0, sizeof d);    d[ss] = 1;    queue<int>q;    q.push(ss);    while (!q.empty())    {        int pre = q.front();        q.pop();        for (int i = s[pre]; ~i; i = edge[i].next)        {            int v = edge[i].v;            if (edge[i].cap > 0 && !d[v])            {                d[v] = d[pre] + 1;                q.push(v);            }        }    }    return d[ee];}int DFS(int x, int exp, int ee){    if (x == ee||!exp) return exp;    int temp,flow=0;    for (int i = nt[x]; ~i ; i = edge[i].next, nt[x] = i)    {        int v = edge[i].v;        if (d[v] == d[x] + 1&&(temp = (DFS(v, min(exp, edge[i].cap), ee))) > 0)        {            edge[i].cap -= temp;            edge[i ^ 1].cap += temp;            flow += temp;            exp -= temp;            if (!exp) break;        }    }    if (!flow) d[x] = 0;    return flow;}int Dinic_flow(int ss, int ee){    int ans = 0;    while (BFS(ss, ee))    {        for (int i = 0; i <= ee; i++) nt[i] = s[i];        ans+= DFS(ss, INF, ee);    }    return ans;}int main(){    int n,m;    while (~scanf("%d%d", &m, &n))    {        init();        int ss,ee,ww;        for(int i=1;i<=m;i++)        {            scanf("%d%d%d",&ss,&ee,&ww);            add(ss,ee,ww);        }        printf("%d\n", Dinic_flow(1, n));    }    return 0;}