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 
    
     #define fi first#define se second#define pii pair
     
      #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 ivorysiusing namespace std;typedef long long int64;typedef unsigned int u32;typedef double db;template
      
       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
       
        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
        
          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();    }}