bzoj1066: [SCOI2007]蜥蜴

最大流,首先建模。 设每个石柱分为出节点和入节点。

1.设一个虚拟源点,与每个初始有蜥蜴的石柱连一条容量为1的边。

2.设一个虚拟汇点,与每个能跳出来的石柱连一条容量为INF的边。

3.每一对距离小于d的点,出节点和入节点连一条容量为INF的边。

4.每个点的入节点和出节点连一条容量为石柱高度的边。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int maxn = 1000 + 10;
const int maxm = 100000 + 10;
const int INF = 0x3f3f3f3f;

int n,m,d,S,T,vid=0,cnt=0,p=0;
int h[maxn];
int id[25][25][2];
int to[maxm],next[maxm],f[maxm];
int dist[maxn],gap[maxn];
char tmp[25];

int abs(int x) {
    return (x > 0 ? x : -x);
}

void add(int u,int v,int F) {
    to[++p] = v; f[p] = F; next[p] = h[u]; h[u] = p;
    to[++p] = u; f[p] = 0; next[p] = h[v]; h[v] = p;
}

int ISAP(int u,int flow) {
    if(u == T) return flow;
    
    int cur = 0,aug,mindist = vid;
        
    for(int i = h[u]; ~i; i = next[i]) {
        if(f[i] && dist[to[i]]+1 == dist[u]) {
            aug = ISAP(to[i],min(f[i],flow-cur));
            f[i] -= aug; 
            f[i^1] += aug; 
            cur += aug;
            if(cur == flow || dist[S] >= vid) 
                return cur;
        }
    }
    
    if(cur == 0) {
        if(!--gap[dist[u]]) {
            dist[S] = vid;
            return cur;
        }
        for(int i = h[u]; ~i; i = next[i]) if(f[i])
            mindist = min(mindist,dist[to[i]]);
        ++gap[dist[u]=mindist+1];
    }
    return cur;
}

int flow() {
    int res = 0;
    gap[0] = vid;
    while(dist[S] < vid) 
        res += ISAP(S,INF);
    return res;
}


void build() {
    memset(h,-1,sizeof(h));
    p = -1;
    
    scanf("%d%d%d",&n,&m,&d);
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++) {
            id[i][j][0] = ++vid;
            id[i][j][1] = ++vid;
        }
    S = ++vid; T = ++vid;
    
    for(int x1 = 1; x1 <= n; x1++)
    for(int y1 = 1; y1 <= m; y1++) {
        for(int x2 = 1; x2 <= n; x2++)
        for(int y2 = 1; y2 <= m; y2++)
            if(abs(x1-x2) + abs(y1-y2) <= d) 
            if(x1 != x2 || y1 != y2) 
                add(id[x1][y1][1],id[x2][y2][0],INF);
        if(x1 - d < 1 || x1 + d > n || y1 - d < 1 || y1 + d > m)
            add(id[x1][y1][1],T,INF); 
    }
    
    for(int i = 1; i <= n; i++) {
        scanf("%s",tmp+1);
        for(int j = 1; j <= m; j++) 
            add(id[i][j][0],id[i][j][1],tmp[j]-'0');
    }
    
    for(int i = 1; i <= n; i++) {
        scanf("%s",tmp+1);
        for(int j = 1; j <= m; j++)
            if(tmp[j] == 'L') {
                add(S,id[i][j][0],1);
                cnt++;
            }
    }
    
}

void solve() {
    printf("%d\n",cnt - flow());
}

int main() {
    build();
    solve();
    return 0;
}
posted @ 2016-04-10 10:03  invoid  阅读(141)  评论(0编辑  收藏