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题目链接:http://poj.org/problem?id=3281
题意:
给出牛n,饮料d还有食物f的数量,每头牛给出喜欢的饮料和食物,最后求出能够满足的牛的数量
解法:源点到食物建边, 由食物到牛建边, 牛到饮料建边 ,饮料到汇点建边 ,求最大流。牛要拆点控制流量为1
全部是有向的边,而且权值全部为1
有2*n+f+d+2个顶点
0表示源点,2*n+f+d+1表示汇点
1到f为食物点,f+1到f+2*n为牛点,f+2*n+1到f+2*n+d为饮料点
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>
#pragma comment (linker, "/STACK:102400000,102400000")
using namespace std;
const int MAXN = 1010;//点数的最大值
const int MAXM = 400010;//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 m, n, f, d, x, y;
int a, b, c;
int main()
{
while (scanf("%d%d%d", &n, &f, &d) != EOF)
{
init();
int tmp = f + 2*n;
int so = 0;//源点
int to = f + n * 2 + d + 1;//汇点
int tot = f + n * 2 + d + 2;
for (int i = 1; i <= f; i++)//源点指向食物
addedge(so, i, 1);
for (int i = 1; i <= d; i++)//饮料指向汇点
addedge(tmp + i , to, 1);
for (int i = 1; i <= n; i++)//牛指向牛
addedge(f + i, f + n + i, 1);
for (int i = 1; i <= n;i++)
{
scanf("%d%d",&x,&y);
while (x--)
{
scanf("%d",&m);
addedge(m, f + i, 1);
}
while (y--)
{
scanf("%d", &m);
addedge(f + n + i, 2*n + f + m, 1);
}
}
int ans = sap(so,to, tot);
printf("%d\n", ans);
}
return 0;
}
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原文地址:http://blog.csdn.net/u014427196/article/details/47340833