[BJOI2012]连连看 BZOJ2661 费用流

题目描述

凡是考智商的题里面总会有这么一种消除游戏。不过现在面对的这关连连看可不是QQ游戏里那种考眼力的游戏。我们的规则是，给出一个闭区间[a,b]中的全部整数，如果其中某两个数x,y（设x>y）的平方差x^2-y^2是一个完全平方数z^2，并且y与z互质，那么就可以将x和y连起来并且将它们一起消除，同时得到x+y点分数。那么过关的要求就是，消除的数对尽可能多的前提下，得到足够的分数。快动手动笔算一算吧。

输入输出格式

只有一行，两个整数，分别表示a,b。

两个数，可以消去的对数，及在此基础上能得到的最大分数。

输入输出样例

1 15

2 34

说明

对于30%的数据，1<=a,b<=100

对于100%的数据，1<=a,b<=1000

将每个点分为in,out；

建边的时候要注意我们消去的是一对数字；

所以建边的时候Inx----Outy,Iny----Outx都要建边；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 100005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;

inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}


ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }



/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/
bool vis[maxn];
int n, m, s, t;
int x, y, f, z;
int dis[maxn], pre[maxn], last[maxn], flow[maxn];
int maxflow, mincost;

struct node {
	int to, nxt, flow, dis;
}edge[maxn << 2];

int head[maxn], cnt;
queue<int>q;

void addedge(int from, int to, int flow, int dis) {
	edge[++cnt].to = to; edge[cnt].flow = flow; edge[cnt].dis = dis;
	edge[cnt].nxt = head[from]; head[from] = cnt;
}

bool spfa(int s, int t) {
	memset(dis, 0x7f, sizeof(dis)); memset(flow, 0x7f, sizeof(flow));
	ms(vis);
	q.push(s); vis[s] = 1; dis[s] = 0; pre[t] = -1;
	while (!q.empty()) {
		int now = q.front(); q.pop(); vis[now] = 0;
		for (int i = head[now]; i != -1; i = edge[i].nxt) {
			if (edge[i].flow > 0 && dis[edge[i].to] > dis[now] + edge[i].dis) {
				dis[edge[i].to] = edge[i].dis + dis[now];
				pre[edge[i].to] = now; last[edge[i].to] = i;
				flow[edge[i].to] = min(flow[now], edge[i].flow);
				if (!vis[edge[i].to]) {
					vis[edge[i].to] = 1; q.push(edge[i].to);
				}
			}
		}
	}
	return pre[t] != -1;
}

void mincost_maxflow() {
	while (spfa(s, t)) {
		int now = t;
		maxflow += flow[t]; mincost += flow[t] * dis[t];
		while (now != s) {
			edge[last[now]].flow -= flow[t];
			edge[last[now] ^ 1].flow += flow[t];
			now = pre[now];
		}
	}
}

bool OK(int x, int y) {
	if (x < y)swap(x, y);
	int Z = x * x - y * y;
	int z = (int)sqrt(Z);
	if (z*z == Z) {
		if (gcd(z, y) == 1) {
			return true;
		}
		else return false;
	}
	else return false;
}

int main()
{
//	ios::sync_with_stdio(0);
	mclr(head, -1); cnt = 1;
	int a, b; a = rd(); b = rd();
	s = 0; t = b + 1;
	int num = b - a + 1;
	for (int i = a; i <= b; i++) {
		addedge(s, i, 1, 0); addedge(i, s, 0, 0);
		addedge(i + b , t, 1, 0); addedge(t, i + b , 0, 0);
	}
	for (int i = a; i <= b; i++) {
		for (int j = i + 1; j <= b; j++) {
			if (OK(i, j)) {
				addedge(i + b, j, 0, i + j); addedge(j, i + b, 1, -(i + j));
				addedge(j + b, i, 0, i + j); addedge(i, j + b, 1, -(i + j));
			}
		}
	}
	mincost_maxflow();
	printf("%d %d\n", maxflow/2, -mincost/2);
	return 0;
}

标签：now,连连看,int,BJOI2012,flow,edge,BZOJ2661,include,define
来源： https://www.cnblogs.com/zxyqzy/p/10353740.html