HDU-1532-Drainage Ditches

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ACM模版

描述

题解

基础的最大流问题，模版题，我用了一个十分成熟的 Dinic 算法模版，用了邻接表优化，代码略微长，但是效率还是不错的。GG

代码

#include <iostream>#include <cstring>#include <cstdio>/* *  Dinic 最大流 O(V^2 * E) *  INIT: ne=2; head[]置为0; addedge()加入所有弧; *  CALL: flow(n, s, t); */#define typec int               //  type of costusing namespace std;const typec inf = 0x3f3f3f3f;   // max of costconst typec MAXE = 410;const typec MAXN = 210;struct edge{    int x, y, nxt;    typec c;} bf[MAXE];int ne, head[MAXN], cur[MAXN], ps[MAXN], dep[MAXN];void addedge(int x, int y, typec c){   //  add an arc(x->y, c);    vertex:0~n-1;    bf[ne].x = x;    bf[ne].y = y;    bf[ne].c = c;    bf[ne].nxt = head[x];    head[x] = ne++;    bf[ne].x = y;    bf[ne].y = x;    bf[ne].c = 0;    bf[ne].nxt = head[y];    head[y] = ne++;    return ;}typec flow(int n, int s, int t){    typec tr, res = 0;    int i, j, k, f, r, top;    while (1)    {        memset(dep, -1, n * sizeof(int));        for (f = dep[ps[0] = s] = 0, r = 1; f != r;)        {            for (i = ps[f++], j = head[i]; j; j = bf[j].nxt)            {                if (bf[j].c && -1 == dep[k = bf[j].y])                {                    dep[k] = dep[i] + 1;                    ps[r++] = k;                    if (k == t)                    {                        f = r;                        break;                    }                }            }        }        if (-1 == dep[t])        {            break;        }        memcpy(cur, head, n * sizeof(int));        for (i = s, top = 0; ;)        {            if (i == t)            {                for (k = 0, tr = inf; k < top; ++k)                {                    if (bf[ps[k]].c < tr)                    {                        tr = bf[ps[f = k]].c;                    }                }                for (k = 0; k < top; ++k)                {                    bf[ps[k]].c -= tr, bf[ps[k]^1].c += tr;                }                res += tr;                i = bf[ps[top = f]].x;            }            for (j = cur[i]; cur[i]; j = cur[i] = bf[cur[i]].nxt)            {                if (bf[j].c && dep[i] + 1 == dep[bf[j].y])                {                    break;                }            }            if (cur[i])            {                ps[top++] = cur[i];                i = bf[cur[i]].y;            }            else            {                if (0 == top)                {                    break;                }                dep[i] = -1;                i = bf[ps[--top]].x;            }        }    }    return res;}int E, N;int main(){    while (cin >> E >> N)    {        ne = 2;        memset(head, 0, sizeof(head));        int s, t;        typec w;        while (E--)        {            scanf("%d%d%d", &s, &t, &w);            addedge(s - 1, t - 1, w);        }        typec ans = flow(N, 0, N - 1);        printf("%d\n", ans);    }    return 0;}

参考

《最大流》

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