[CCPC2019秦皇岛] E. Escape

[CCPC2019秦皇岛E] Escape

Link

https://codeforces.com/gym/102361/problem/E

Solution

观察到性质若干然后建图跑最大流即可。

我的 ISAP 被卡了，换成 Dinic 却过了。

#include <bits/stdc++.h>
using namespace std;

const int maxn = 200005;
const int inf = 1e+9;

#define reset(x) memset(x,0,sizeof x)

int dis[maxn], ans, cnt = 1, s, t, pre[maxn * 10], nxt[maxn * 10], h[maxn], v[maxn * 10];
std::queue<int> q;
void make(int x, int y, int z) {
    pre[++cnt] = y, nxt[cnt] = h[x], h[x] = cnt, v[cnt] = z;
    pre[++cnt] = x, nxt[cnt] = h[y], h[y] = cnt;
}
bool bfs() {
    memset(dis, 0, sizeof dis);
    q.push(s), dis[s] = 1;
    while (!q.empty()) {
        int x = q.front();
        q.pop();
        for (int i = h[x]; i; i = nxt[i])
            if (!dis[pre[i]] && v[i])
                dis[pre[i]] = dis[x] + 1, q.push(pre[i]);
    }
    return dis[t];
}
int dfs(int x, int flow) {
    if (x == t || !flow)
        return flow;
    int f = flow;
    for (int i = h[x]; i; i = nxt[i])
        if (v[i] && dis[pre[i]] > dis[x]) {
            int y = dfs(pre[i], min(v[i], f));
            f -= y, v[i] -= y, v[i ^ 1] += y;
            if (!f)
                return flow;
        }
    if (f == flow)
        dis[x] = -1;
    return flow - f;
}

int T,n,m,a,b,p[105],e[105];
char c[105][105];

int calch(int i,int j)
{
    return 3+((i-1)*m+j-1)*2;
}
int calcv(int i,int j)
{
    return 4+((i-1)*m+j-1)*2;
}
int calcx(int i,int j)
{
    return 3+2*n*m+((i-1)*m+j-1);
}
int calcy(int i,int j)
{
    return 3+3*n*m+((i-1)*m+j-1);
}

signed main()
{
    ios::sync_with_stdio(false);
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d%d",&n,&m,&a,&b);
        reset(dis); reset(pre); reset(nxt); reset(h); reset(v); cnt=1;
        //cout<<calch(n,m)<<" "<<calcv(n,m)<<" "<<calcx(n,m)<<" "<<calcy(n,m)<<endl;
        for(int i=1;i<=n;i++) scanf("%s",c[i]+1);
        for(int i=1;i<=a;i++) scanf("%d",&p[i]);
        for(int i=1;i<=b;i++) scanf("%d",&e[i]);
        for(int i=1;i<=n-1;i++)
            for(int j=1;j<=m;j++)
                if(c[i][j]=='0' && c[i+1][j]=='0')
                    make(calcv(i,j),calcv(i+1,j),1),
                    make(calcv(i+1,j),calcv(i,j),1);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m-1;j++)
                if(c[i][j]=='0' && c[i][j+1]=='0')
                    make(calch(i,j),calch(i,j+1),1),
                    make(calch(i,j+1),calch(i,j),1);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                if(c[i][j]=='0')
                    make(calch(i,j),calcv(i,j),1),
                    make(calcv(i,j),calch(i,j),1);
        for(int i=1;i<=a;i++) if(c[1][p[i]]=='0') make(1,calcv(1,p[i]),1);
        for(int i=1;i<=b;i++) if(c[n][e[i]]=='0') make(calcv(n,e[i]),2,1);
        s=1;t=2;
        ans = 0;
        for (; bfs(); ans += dfs(s, inf));
        if(ans == a) printf("Yes\n");
        else printf("No\n");
    }
}

原文地址：https://www.cnblogs.com/mollnn/p/11731500.html