UVA 10779 Collectors Problem 网络流+建图

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题目链接：点击打开链接

题意：白书P370

思路：

因为问的是最后贴纸总数，那么就设最后的贴纸总数是网络流的答案。

首先我们模拟贴纸的流动过程：

Bob 的 某种贴纸a -> 给一个没有a贴纸的人Peo -> 还给Bob一个Peo的某张重复贴纸 -> 这张贴纸可以算作答案了

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<vector>
#include <queue>
using namespace std;
#define ll int
const int MAXN = 100010;//点数的最大值
const int MAXM = 400010;//边数的最大值
#define N MAXN
#define M MAXM
const int INF = 0x3f3f3f3f;
#define inf INF
struct Edge{
	ll from, to, cap, nex;
}edge[M*2];//注意这个一定要够大 不然会re 还有反向弧

ll head[N], edgenum;
void add(ll u, ll v, ll cap, ll rw = 0){ //如果是有向边则：add(u,v,cap); 如果是无向边则：add(u,v,cap,cap);
	Edge E = { u, v, cap, head[u]};
	edge[ edgenum ] = E;
	head[u] = edgenum ++;

	Edge E2= { v, u, rw,  head[v]};
	edge[ edgenum ] = E2;
	head[v] = edgenum ++;
}
ll sign[N];
bool BFS(ll from, ll to){
	memset(sign, -1, sizeof(sign));
	sign[from] = 0;

	queue<ll>q;
	q.push(from);
	while( !q.empty() ){
		ll u = q.front(); q.pop();
		for(ll i = head[u]; i!=-1; i = edge[i].nex)
		{
			ll v = edge[i].to;
			if(sign[v]==-1 && edge[i].cap)
			{
				sign[v] = sign[u] + 1, q.push(v);
				if(sign[to] != -1)return true;
			}
		}
	}
	return false;
}
ll Stack[N], top, cur[N];
ll Dinic(ll from, ll to){
	ll ans = 0;
	while( BFS(from, to) )
	{
		memcpy(cur, head, sizeof(head));
		ll u = from;      top = 0;
		while(1)
		{
			if(u == to)
			{
				ll flow = inf, loc;//loc 表示 Stack 中 cap 最小的边
				for(ll i = 0; i < top; i++)
					if(flow > edge[ Stack[i] ].cap)
					{
						flow = edge[Stack[i]].cap;
						loc = i;
					}

					for(ll i = 0; i < top; i++)
					{
						edge[ Stack[i] ].cap -= flow;
						edge[Stack[i]^1].cap += flow;
					}
					ans += flow;
					top = loc;
					u = edge[Stack[top]].from;
			}
			for(ll i = cur[u]; i!=-1; cur[u] = i = edge[i].nex)//cur[u] 表示u所在能增广的边的下标
				if(edge[i].cap && (sign[u] + 1 == sign[ edge[i].to ]))break;
			if(cur[u] != -1)
			{
				Stack[top++] = cur[u];
				u = edge[ cur[u] ].to;
			}
			else
			{
				if( top == 0 )break;
				sign[u] = -1;
				u = edge[ Stack[--top] ].from;
			}
		}
	}
	return ans;
}
void init(){memset(head,-1,sizeof head); edgenum = 0;}

int n, m, k, s, t;
int a[30];
void go(){
    int x;
    scanf("%d", &x);
    memset(a, 0, sizeof a);
    while(x--){
        int y; scanf("%d",&y); a[y]++;
    }
}
void solve(){
	init();
    scanf("%d %d", &n, &m);
    s = 1, t = n+m+1;
    go();
    for(int i = 1; i <= m; i++) if(a[i])
        add(s, n+i, a[i]);
    for(int i = 2; i <= n; i++)
    {
        go();
        for(int j = 1; j <= m; j++)
        if(a[j]-1 > 0)
            add(i, n+j, a[j] -1);
        else if(a[j]==0)
            add(n+j, i, 1);
    }
    for(int i = 1; i <= m; i++)
        add(n+i, t, 1);
    printf("%d\n", Dinic(s,t));
}
int main(){
    int T, Cas = 1; scanf("%d",&T);
	while(T--){
        printf("Case #%d: ", Cas++);
		solve();
    }
	return 0;
}

UVA 10779 Collectors Problem 网络流+建图

标签：style   blog   http   color   io   os   ar   for   2014   

原文地址：http://blog.csdn.net/qq574857122/article/details/39300185