版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/pengwill97/article/details/82853291
题意
在一个r行c列的网格地图中有一些高度不同的石柱,一些石柱上站着一些蜥蜴,你的任务是让尽量多的蜥蜴逃到边界外。
每行每列中相邻石柱的距离为1,蜥蜴的跳跃距离是d,即蜥蜴可以跳到平面距离不超过d的任何一个石柱上。石柱都不稳定,每次当蜥蜴跳跃时,所离开的石柱高度减1(如果仍然落在地图内部,则到达的石柱高度不变),如果该石柱原来高度为1,则蜥蜴离开后消失。以后其他蜥蜴不能落脚。任何时刻不能有两只蜥蜴在同一个石柱上。
题解
最大流。
建立源点S,S向每个蜥蜴的位置连一条边,容量为1。石柱的次数限制,用拆点的方法。及把一个点拆成2个点,前点和后点,前点向后点连边,容量为石柱的次数。每个石柱可以互相跳到的,连边,容量为INF。如果石柱能跳到外面,那么石柱向汇点T连边,容量为INF。
最后跑最大流即可。
代码
//
// Created by pengwill on 2018/9/26.
//
#include <bits/stdc++.h>
using namespace std;
const int nmax = 1e6 + 7;
const int INF = 0x3f3f3f3f;
typedef pair<int, int> pii;
vector<pii> pos;
char mp[100][100];
char lizerd[100][100];
int n, m, k;
inline int baseid(int x, int y) {
return (x - 1) * m + y;
}
inline int exid(int x, int y ) {
return n * m + baseid(x, y);
}
inline bool legal(int x, int y) {
if(x < 1 || x > n || y < 1 || y > m) return false;
else return true;
}
struct Dinic {
int head[nmax], cur[nmax], d[nmax];
bool vis[nmax],iscut[nmax];
int tot, n, m, s, t;
struct edge {
int nxt, to, w, cap, flow;
} e[nmax<<1];
void init(int n) {
this->n = n;
this->tot = 0;
memset(head, -1, sizeof head);
memset(iscut,0,sizeof iscut);
}
void add_edge(int u, int v, int c) {
e[tot].to = v, e[tot].cap = c, e[tot].flow = 0;
e[tot].nxt = head[u];
head[u] = tot++;
e[tot].to = u, e[tot].cap = c, e[tot].flow = c;
e[tot].nxt = head[v];
head[v] = tot++;
}
bool BFS() {
memset(vis, 0, sizeof(vis));
queue<int>Q;
vis[s] = 1; d[s] = 0;
Q.push(s);
while (!Q.empty()) {
int u = Q.front(); Q.pop();
for (int i = head[u]; i != -1; i = e[i].nxt) {
int v = e[i].to;
if (!vis[v] && e[i].cap > e[i].flow) {
vis[v] = 1;
d[v] = d[u] + 1;
Q.push(v);
}
}
}
return vis[t];
}
int DFS(int x, int a) {
if (x == t || a == 0) return a;
int Flow = 0, f;
for (int& i = cur[x]; i != -1; i = e[i].nxt) {
int v = e[i].to;
if (d[v] == d[x] + 1 && (f = DFS(v, min(a, e[i].cap - e[i].flow))) > 0) {
Flow += f;
e[i].flow += f;
e[i ^ 1].flow -= f;
a -= f;
if (a == 0) break;
}
}
return Flow;
}
int Maxflow(int s, int t) {
this->s = s, this->t = t;
int Flow = 0;
while (BFS()) {
for (int i = 0; i <= n; i++) cur[i] = head[i];
while(int once = DFS(s, INF)) Flow += once;
}
return Flow;
}
void get_cut(int u){
iscut[u] = true;
for (int i = head[u]; i != -1; i = e[i].nxt) {
int v = e[i].to;
if (!iscut[v] && e[i].cap > e[i].flow)
get_cut(v);
}
return;
}
} dinic;
int main() {
scanf("%d %d %d", &n, &m, &k);
int cnt = 0;
int s = 0, t = n * m * 16 + 1;
dinic.init(t);
for(int i = 1; i <= n; ++i)
scanf("%s", mp[i] + 1);
for(int i = 1; i <= n; ++i)
scanf("%s", lizerd[i] + 1);
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= m; ++j) {
mp[i][j] -= '0';
if(mp[i][j] > 0) {
dinic.add_edge(baseid(i, j), exid(i, j), (int)mp[i][j]);
// printf("selfedge (%d, %d, %d)\n", i, j, (int)mp[i][j]);
if(i <= k || j <= k || abs(n - i) < k || abs(m - j) < k)
dinic.add_edge(exid(i, j), t, INF);
// printf("toSink (%d, %d)\n", i, j);
}
if(lizerd[i][j] == 'L')
dinic.add_edge(s, baseid(i, j), 1), cnt ++;
// printf("Sorce (%d, %d)\n", i, j);
}
for(int i = 1; i <= n; ++i) {
for(int j = 1; j <= m; ++j) {
if(mp[i][j] > 0) {
for(int xx = -k; xx <= k; ++xx) {
for(int yy = -k; yy <= k; ++yy) {
int nx = i + xx;
int ny = j + yy;
int tmp = abs(xx) + abs(yy);
if( tmp <= k && tmp != 0 )
if(legal(nx, ny) && mp[nx][ny] > 0) {
dinic.add_edge(exid(i, j), baseid(nx, ny), INF);
// printf("addedge (%d, %d) - (%d, %d)\n", i, j, nx, ny);
}
}
}
}
}
}
int mxflee = dinic.Maxflow(s ,t);
printf("%d\n", cnt - mxflee);
}