POJ-1273 Drainage Ditches【最大流】

题目链接：http://poj.org/problem?id=1273

题目大意：

有一个水塘，一个大海，要把水塘里的水尽可能多的排到大海里。当然，要通过一些水渠，这些水渠有一定的容量限制。

问最多能有多少水能流到大海里。

解题思路：

最大流的入门题。看了一上午就做了出来，很简单其实。。。用的BFS和Ford_Fulkerson算法。网上资料很多，大概看看应该就理解了。网络流开始还是很是不难的。建图是关键，要学会一定的建图方法。。。

代码如下：

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<queue>#include<cmath>using namespace std;#define N 210#define MAX 1 << 25int network[N][N];bool visit[N];int que[N], head, tail; //广搜队列int pre[N]; //祖先（用于更新残余网络）int flow[N]; //可行流量int n, m;int BFS() //增广路{head = tail = 0;for(int i = 1; i <= m; ++i)pre[i] = -1;for(int i = 1; i <= m; ++i)flow[i] = MAX;memset(visit, false, sizeof(visit));visit[1] = true;que[tail++] = 1;while(head < tail){int cur = que[head++];if(cur == m)break;for(int i = 2; i <= m; ++i){if(!visit[i] && network[cur][i]) //增广路{visit[i] = true;flow[i] = min(flow[cur], network[cur][i]);pre[i] = cur;que[tail++] = i;}}}if(!visit[m])return -1;elsereturn flow[m];}int Ford_Fulkerson(){int maxflow = 0, nowflow;while((nowflow = BFS()) != -1){maxflow += nowflow;int temp = m, res;while(temp != 1) //更新残余网络{res = pre[temp];network[res][temp] -= nowflow;network[temp][res] += nowflow;temp = res;}}return maxflow;}int main(){int start, end, water;while(~scanf("%d%d", &n, &m)){memset(network, 0, sizeof(network));for(int i = 0; i < n; ++i){scanf("%d%d%d", &start, &end, &water); //重边叠加~。~network[start][end] += water;}printf("%d\n", Ford_Fulkerson());}return 0;}

SAP实现代码如下：

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<cmath>using namespace std;#define N 210#define MAX 1<<28#define CLR(arr, what) memset(arr, what, sizeof(arr))int maze[N][N];int pre[N], cur[N];int gap[N];int dis[N];int source, sink, nodenum;int SAP(int s, int t, int n){CLR(gap, 0); CLR(cur, 0); CLR(dis, 0);int u = pre[s] = s, maxflow = 0, aug = MAX, v;gap[0] = n;while(dis[s] < n){bool flag = false;for(v = cur[u]; v <= n; ++v) //寻找允许弧{if(maze[u][v] > 0 && dis[u] == dis[v] + 1){flag = true;break;}}if(flag) //找到允许弧{pre[v] = u;cur[u] = v;aug = min(aug, maze[u][v]);u = v;if(v == t) //找到完整增广路{maxflow += aug;for(v = t; v != s; v = pre[v]) //更新残留网络{maze[pre[v]][v] -= aug; //正向边maze[v][pre[v]] += aug; //反向边}aug = MAX, u = s; //重新从源点寻找}}else //找不到允许弧{int mindis = n;for(v = 1; v <= n; ++v) //重新标号{if(maze[u][v] && mindis > dis[v]){cur[u] = v;mindis = dis[v];}}if(--gap[dis[u]] == 0) //更新断层 + 判断是否断层（间隙优化）break;gap[dis[u] = mindis + 1]++; //更新断层u = pre[u]; //当前弧优化}}return maxflow;}int main(){int n, m;int start, end, cost;int answer;while(~scanf("%d%d", &n, &m)){CLR(maze, 0);source = 1, sink = m, nodenum = m;for(int i = 0; i < n; ++i){scanf("%d%d%d", &start, &end, &cost);maze[start][end] += cost;}answer = SAP(source, sink, nodenum);printf("%d\n", answer);}return 0;}