bzoj3175 [Tjoi2013]攻击装置 二分图匹配

Description

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给定一个01矩阵，其中你可以在0的位置放置攻击装置。每一个攻击装置（x,y）都可以按照“日”字攻击其周围的 8个位置（x-1,y-2）,(x-2,y-1),(x+1,y-2),(x+2,y-1),(x-1,y+2),(x-2,y+1), (x+1,y+2),(x+2,y+1)
求在装置互不攻击的情况下，最多可以放置多少个装置。

100%数据 N<=200

Solution

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总点数-二分图匹配。之前做过poj类似的题算是复习了
根据点横纵坐标分奇偶点，然后连边即可。加上当前弧优化跑得巨快

Code

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#include <stdio.h>
#include <string.h>
#include <queue>
#define rep(i,st,ed) for (int i=st;i<=ed;++i)
#define fill(x,t) memset(x,t,sizeof(x))
#define min(x,y) ((x)<(y)?(x):(y))
const int INF=0x3f3f3f3f;
const int N=80505;
const int E=640005;
const int L=205;
struct edge{int x,y,w,next;}e[E];
std:: queue<int> queue;
int dx[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,2},{1,-2},{2,1},{2,-1}};
int cur[N],ls[N],edCnt=1;
int dis[N];
char rc[L][L];
int read() {
    int x=0,v=1; char ch=getchar();
    for (;ch<'0'||ch>'9';v=(ch=='-')?(-1):v,ch=getchar());
    for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());
    return x*v;
}
void addEdge(int x,int y,int w) {
    e[++edCnt]=(edge){x,y,w,ls[x]}; ls[x]=edCnt;
    e[++edCnt]=(edge){y,x,0,ls[y]}; ls[y]=edCnt;
}
bool bfs(int st,int ed) {
    while (!queue.empty()) queue.pop();
    queue.push(st);
    fill(dis,-1);
    dis[st]=1;
    while (!queue.empty()) {
        int now=queue.front(); queue.pop();
        for (int i=ls[now];i;i=e[i].next) {
            if (e[i].w>0&&dis[e[i].y]==-1) {
                queue.push(e[i].y);
                dis[e[i].y]=dis[now]+1;
                if (e[i].y==ed) return 1;
            }
        }
    }
    return 0;
}
int dfs(int now,int ed,int mn) {
    if (now==ed||!mn) return mn;
    int ret=0;
    for (int &i=cur[now];i;i=e[i].next) {
        if (e[i].w>0&&dis[now]+1==dis[e[i].y]) {
            int d=dfs(e[i].y,ed,min(e[i].w,mn-ret));
            e[i].w-=d; e[i^1].w+=d;
            ret+=d;
            if (ret==mn) break;
        }
    }
    return ret;
}
int dinic(int st,int ed) {
    int ret=0;
    while (bfs(st,ed)) {
        rep(i,st,ed) cur[i]=ls[i];
        ret+=dfs(st,ed,INF);
    }
    return ret;
}
int main(void) {
    int n; scanf("%d",&n);
    int cnt=0;
    rep(i,1,n) scanf("%s", rc[i]+1);
    rep(i,1,n) rep(j,1,n) {
        if (rc[i][j]=='0') cnt++;
        if ((i^j)&1&&rc[i][j]=='0') {
            addEdge(0,(i-1)*n+j,1);
            rep(k,0,7) {
                int ti=i+dx[k][0],tj=j+dx[k][1];
                if (ti>0&&ti<=n&&tj>0&&tj<=n&&rc[ti][tj]=='0') {
                    addEdge((i-1)*n+j,(ti-1)*n+tj,1);
                }
            }
        } else if (rc[i][j]=='0') addEdge((i-1)*n+j,n*n+1,1);
    }
    int mxFlow=dinic(0,n*n+1);
    printf("%d\n",cnt-mxFlow);
    return 0;
}