标签:blog os io for div ar amp type
建模的思路大概是这样的,把房间当做点,门当做是边,如果从房间A能到房间B中间有一个门,如果锁在A这边那么A->B容量就是INF,B->A的容量就是1。
攻击者如果在A这边的话显然就算你锁了门也是没有意义的,在B这边如果锁上是有意义的,所以算1个门,然后就很简单了,建立源点到所有攻击者点的边,容量为INF,汇点就是要保护的那个房间。求最小割即可。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <string>
#include <iostream>
#include <map>
#include <cstdlib>
#include <list>
#include <set>
#include <queue>
#include <stack>
using namespace std;
typedef long long LL;
const int maxn = 30;
const int INF = INT_MAX / 3;
int cap[maxn][maxn],n,s,t;
int level[maxn],q[maxn],qs,qe;
char buf[128];
bool bfs() {
qs = qe = 0;
q[qe++] = s;
memset(level,0,sizeof(level));
level[s] = 1;
while(qs < qe) {
int now = q[qs++];
for(int i = 0;i <= n;i++) if(cap[now][i] && level[i] == 0) {
level[i] = level[now] + 1; q[qe++] = i;
}
}
return level[t];
}
int dfs(int now,int alpha) {
if(now == t) return alpha;
int sum = 0;
for(int i = 0;i <= n && alpha;i++) {
if(cap[now][i] && level[i] == level[now] + 1) {
int ret = dfs(i,min(alpha,cap[now][i]));
sum += ret; alpha -= ret;
cap[now][i] -= ret; cap[i][now] += ret;
}
}
if(sum == 0) level[now] = -1;
return sum;
}
void solve() {
int ans = 0;
while(bfs()) ans += dfs(s,INT_MAX);
if(ans >= INF) puts("PANIC ROOM BREACH");
else printf("%d\n",ans);
}
int main() {
int T; scanf("%d",&T);
while(T--) {
memset(cap,0,sizeof(cap));
scanf("%d%d",&n,&t);
s = n;
for(int i = 0;i < n;i++) {
int m;
scanf("%s%d",buf,&m);
for(int j = 0;j < m;j++) {
int tmp; scanf("%d",&tmp);
cap[i][tmp] = INF;
cap[tmp][i]++;
}
if(buf[0] == ‘I‘) cap[s][i] = INF;
}
solve();
}
return 0;
}
POJ 3084 Panic Room 求最小割,布布扣,bubuko.com
标签:blog os io for div ar amp type
原文地址:http://www.cnblogs.com/rolight/p/3873642.html