[BZOJ1001]狼抓兔子

2020-03-01
BZOJ

题意

问下图的最小割,S左上,T右下,$n,m\leq 1000$

题解

直接Dinic怕是不太行(也有大仙过了)

平面对偶图,求最短路即可

注意n=1,m=1时特判

调试记录

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#include <cstdio>
#include <queue>
#include <algorithm>
#include <cstring>
#define ID(i, j) ((i) * m + (j))
const int maxn = 2e6 + 5;
const int S = 1e6 + 1;
const int T = 1e6 + 2;
using namespace std;
struct E{
int to, nxt, l, f;
}e[maxn << 2];
int head[maxn], tot = 0;
void addedge(int u, int v, int l){
e[++tot].to = v, e[tot].nxt = head[u], e[tot].l = l;
head[u] = tot;
e[tot].f = u;
}
priority_queue <pair<int, int> > q;
int dis[maxn]; bool vis[maxn];
void Dijkstra(int S){
q.push(make_pair(0, S));
memset(vis, 0, sizeof vis);
memset(dis, 0x3f, sizeof dis); dis[S] = 0;
while (!q.empty()){
int cur = q.top().second; q.pop();
if (vis[cur]) continue; vis[cur] = 1;
for (int i = head[cur]; i; i = e[i].nxt){
int v = e[i].to;
if (dis[v] > dis[cur] + e[i].l){
dis[v] = dis[cur] + e[i].l;
q.push(make_pair(-dis[v], v));
}
}
}
}
int n, m;
int main(){
// freopen("1.in", "r", stdin);
scanf("%d%d", &n, &m); --n, --m;
if (n == 0){
int ans = 0x3f3f3f3f;
for (int k, i = 0; i < m; i++){
scanf("%d", &k); ans = min(ans, k);
}
printf("%d\n", ans);
return 0;
}
if (m == 0){
int ans = 0x3f3f3f3f;
for (int k, i = 0; i < n; i++){
scanf("%d", &k); ans = min(ans, k);
}
printf("%d\n", ans);
return 0;
}
for (int i = 0; i <= n; i++){
for (int k, j = 0; j < m; j++){
scanf("%d", &k);
if (i == 0) addedge(ID(i, j) << 1, T, k);
else if (i == n) addedge(S, ID(i - 1, j) << 1 | 1, k);
else addedge(ID(i, j) << 1, ID(i - 1, j) << 1 | 1, k), addedge(ID(i - 1, j) << 1 | 1, ID(i, j) << 1, k);
}
}
for (int i = 0; i < n; i++){
for (int k, j = 0; j <= m; j++){
scanf("%d", &k);
if (j == 0) addedge(S, ID(i, j) << 1 | 1, k);
else if (j == m) addedge(ID(i, j - 1) << 1, T, k);
else addedge(ID(i, j - 1) << 1, ID(i, j) << 1 | 1, k), addedge(ID(i, j) << 1 | 1, ID(i, j - 1) << 1, k);
}
}
for (int i = 0; i < n; i++){
for (int k, j = 0; j < m; j++){
scanf("%d", &k);
addedge(ID(i, j) << 1 | 1, ID(i, j) << 1, k);
addedge(ID(i, j) << 1, ID(i, j) << 1 | 1, k);
}
}
// for (int i = 1; i <= 20; i++) printf("%d -> %d (%d)\n", e[i].f, e[i].to, e[i].l);
Dijkstra(S);
printf("%d\n", dis[T]);
return 0;
}