最大流入门题传送门:POJ 1273 Drainage Ditches
下面是ISAP + 当前弧优化 + GAP优化的代码:
#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; }