LOJ#2239. 「CQOI2014」危桥

就是先把每条边正着连一条容量为2的边，反着连一条容量为2的边

显然如果只有一个人走的话，答案就是一个源点往起点连一条容量为次数×2的边，终点往汇点连一个次数×2的边，跑最大流看是否满流即可

两个人的话由于两个人的路径可能相交，有可能从\(a_1\)走到了\(b_2\)

统计一遍 \(a_1,b_{1}\)为源点，\(a_{2},b_{2}\)为汇点的情况

再统计一遍\(a_{1},b_{2}\)为源点，\(a_{2},b_{1}\)为汇点的情况

这两种都合法的话才能证明可以走到

#include <bits/stdc++.h>

#define fi first

#define se second

#define pii pair<int,int>

#define mp make_pair

#define pb push_back

#define space putchar(' ')

#define enter putchar('\n')

#define eps 1e-10

#define ba 47

#define MAXN 1005

//#define ivorysi

using namespace std;

typedef long long int64;

typedef unsigned int u32;

typedef double db;

template<class T>

void read(T &res) {

    res = 0;T f = 1;char c = getchar();

    while(c < '0' || c > '9') {

	if(c == '-') f = -1;

	c = getchar();

    }

    while(c >= '0' && c <= '9') {

	res = res * 10 +c - '0';

	c = getchar();

    }

    res *= f;

}

template<class T>

void out(T x) {

    if(x < 0) {x = -x;putchar('-');}

    if(x >= 10) {

	out(x / 10);

    }

    putchar('0' + x % 10);

}

int N;

int a[3],b[3];

int S,T,dis[55];

char g[55][55];

struct node {

    int to,next,cap;

}E[100005];

int head[55],sumE = 1;

void add(int u,int v,int c) {

    E[++sumE].to = v;

    E[sumE].next = head[u];

    E[sumE].cap = c;

    head[u] = sumE;

}

void addtwo(int u,int v,int c) {

    add(u,v,c);add(v,u,0);

}

queue<int> Q;

bool BFS() {

    Q.push(S);

    memset(dis,0,sizeof(dis));

    dis[S] = 1;

    while(!Q.empty()) Q.pop();

    Q.push(S);

    while(!Q.empty()) {

	int u = Q.front();Q.pop();

	if(u == T) return true;

	for(int i = head[u] ; i;  i = E[i].next) {

	    int v = E[i].to;

	    if(E[i].cap > 0 && !dis[v]) {

		dis[v] = dis[u] + 1;

		if(v == T) return true;

		Q.push(v);

	    }

	}

    }

    return dis[T] != 0;

}

int dfs(int u,int aug) {

    if(u == T) return aug;

    int flow = 0;

    for(int i = head[u] ; i; i = E[i].next) {

	int v = E[i].to;

	if(dis[v] == dis[u] + 1) {

	    int t = dfs(v,min(aug - flow,E[i].cap));

	    flow += t;

	    E[i].cap -= t;

	    E[i ^ 1].cap += t;

	    if(flow == aug) return flow;

	}

    }

    return flow;

}

int Dinic() {

    int res = 0;

    while(BFS()) {

	while(int d = dfs(S,1e9)) {

	    res += d;

	}

    }

    return res;

}

void create() {

    sumE = 1;memset(head,0,sizeof(head));

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

	for(int j = 1 ; j <= N ; ++j) {

	    if(g[i][j] == 'N') addtwo(i,j,1e9);

	    else if(g[i][j] == 'O') addtwo(i,j,2);

	}

    }

}

bool Process() {

    create();

    addtwo(S,a[0],2 * a[2]);

    addtwo(S,b[0],2 * b[2]);

    addtwo(a[1],T,2 * a[2]);

    addtwo(b[1],T,2 * b[2]);

    return Dinic() >= 2 * (a[2] + b[2]);

}

void Solve() {

    for(int i = 0 ; i < 3 ; ++i) read(a[i]);

    ++a[0];++a[1];

    for(int i = 0 ; i < 3 ; ++i) read(b[i]);

    ++b[0];++b[1];

    for(int i = 1 ; i <= N ; ++i) scanf("%s",g[i] + 1);

    S = N + 1;T = N + 2;

    bool f = 1;

    f &= Process();

    swap(b[0],b[1]);

    f &= Process();

    if(f) puts("Yes");

    else puts("No");

}

int main(){

#ifdef ivorysi

    freopen("f1.in","r",stdin);

#endif

    while(scanf("%d",&N) != EOF) {

	Solve();

    }

}