hdu 4292 Food【拆点网络流】

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4292

解法：拆人

代码：

#include <stdio.h>
#include <ctime>
#include <math.h>
#include <limits.h>
#include <complex>
#include <string>
#include <functional>
#include <iterator>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <bitset>
#include <sstream>
#include <iomanip>
#include <fstream>
#include <iostream>
#include <ctime>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <time.h>
#include <ctype.h>
#include <string.h>
#include <assert.h>  

using namespace std;

const int MAXN = 50010;//点数的最大值
const int MAXM = 90001000;//边数的最大值
const int INF = 0x3f3f3f3f;

struct Edge
{
    int to, next, cap, flow;
}edge[MAXM];//注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN], dep[MAXN], pre[MAXN], cur[MAXN];
void init()
{
    tol = 0;
    memset(head, -1, sizeof(head));
}
//加边，单向图三个参数，双向图四个参数
void addedge(int u, int v, int w, int rw = 0)
{
    edge[tol].to = v; edge[tol].cap = w; edge[tol].next = head[u];
    edge[tol].flow = 0; head[u] = tol++;
    edge[tol].to = u; edge[tol].cap = rw; edge[tol].next = head[v];
    edge[tol].flow = 0; head[v] = tol++;
}
//输入参数：起点、终点、点的总数
//点的编号没有影响，只要输入点的总数
int sap(int start, int end, int N)
{
    memset(gap, 0, sizeof(gap));
    memset(dep, 0, sizeof(dep));
    memcpy(cur, head, sizeof(head));
    int u = start;
    pre[u] = -1;
    gap[0] = N;
    int ans = 0;
    while (dep[start] < N)
    {
        if (u == end)
        {
            int Min = INF;
            for (int i = pre[u]; i != -1; i = pre[edge[i ^ 1].to])
                if (Min > edge[i].cap - edge[i].flow)
                    Min = edge[i].cap - edge[i].flow;
            for (int i = pre[u]; i != -1; i = pre[edge[i ^ 1].to])
            {
                edge[i].flow += Min;
                edge[i ^ 1].flow -= Min;
            }
            u = start;
            ans += Min;
            continue;
        }
        bool flag = false;
        int v;
        for (int i = cur[u]; i != -1; i = edge[i].next)
        {
            v = edge[i].to;
            if (edge[i].cap - edge[i].flow && dep[v] + 1 == dep[u])
            {
                flag = true;
                cur[u] = pre[v] = i;
                break;
            }
        }
        if (flag)
        {
            u = v;
            continue;
        }
        int Min = N;
        for (int i = head[u]; i != -1; i = edge[i].next)
            if (edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
            {
                Min = dep[edge[i].to];
                cur[u] = i;
            }
        gap[dep[u]]--;
        if (!gap[dep[u]])return ans;
        dep[u] = Min + 1;
        gap[dep[u]]++;
        if (u != start) u = edge[pre[u] ^ 1].to;
    }
    return ans;
}

int n, f, d;
char s[300];
int F[300], D[300];

int main()
{
    while (~scanf("%d%d%d", &n, &f, &d))
    {
        init();
        for (int i = 1; i <= f; i++) scanf("%d", &F[i]);
        for (int i = 1; i <= d; i++) scanf("%d", &D[i]);

        for (int i = 1; i <= f; i++) addedge(0, i, F[i]);
        for (int i = 1; i <= d; i++) addedge(f + n + n + i, f + n + n + d + 1, D[i]);
        for (int i = 1; i <= n; i++) addedge(f + i, f + n + i, 1);

        for (int i = 1; i <= n; i++)
        {

            scanf("%s", s);
            for (int k = 0; k < f; k++)//食物
            {
                if (s[k] == ‘Y‘)
                    addedge(k + 1, f + i , 1);
            }

        }
        for (int i = 1; i <= n; i++)
        {
            scanf("%s", s);
            for (int k = 0; k < d; k++)//水
            {
                if (s[k] == ‘Y‘)
                    addedge(i + f + n, f + n + n + k + 1, 1);
            }
        }
        int ans = sap(0, f + d + n + n + 1, f + d + n + n + 2);
        printf("%d\n",ans);
    }
    return 0;
}

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