标签:pleasant sheep and b hdu 3046 最小割
4 6 1 0 0 1 0 0 0 1 1 0 0 0 2 0 0 0 0 0 0 2 0 1 1 0
Case 1: 4
题意:一个矩阵,1表示羊,2表示狼,然后让把羊和狼隔开,问需要最小的栅栏数。
思路:要把羊和狼隔开,就是去最小割把他们分开,每个格子和周围四个格子建流量为1的边,s向狼建INF的边,羊向t建INF的边。
代码:
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <string> #include <map> #include <stack> #include <vector> #include <set> #include <queue> #pragma comment (linker,"/STACK:102400000,102400000") #define mod 1000000009 #define INF 0x3f3f3f3f #define pi acos(-1.0) #define eps 1e-6 #define lson rt<<1,l,mid #define rson rt<<1|1,mid+1,r #define FRE(i,a,b) for(i = a; i <= b; i++) #define FREE(i,a,b) for(i = a; i >= b; i--) #define FRL(i,a,b) for(i = a; i < b; i++) #define FRLL(i,a,b) for(i = a; i > b; i--) #define mem(t, v) memset ((t) , v, sizeof(t)) #define sf(n) scanf("%d", &n) #define sff(a,b) scanf("%d %d", &a, &b) #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c) #define pf printf #define DBG pf("Hi\n") typedef long long ll; using namespace std; const int MAXM = 200010; const int MAXN = 43000; int n,m,N; int mp[205][205]; struct Edge { int to,next,cap,flow; }edge[MAXM]; int tol; int head[MAXN]; int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN]; void init(int n) { tol=0;N=n; memset(head,-1,sizeof(head)); } //加边,单向图三个参数,双向图四个参数 void addedge(int u,int v,int w,int rw=0) { edge[tol].to=v; edge[tol].cap=w; edge[tol].next=head[u]; edge[tol].flow=0; head[u]=tol++; edge[tol].to=u; edge[tol].cap=rw; edge[tol].next=head[v]; edge[tol].flow=0; head[v]=tol++; } //输入参数:起点,终点,点的总数 //点的编号没有影响,只要输入点的总数 int sap(int start,int end,int N) { memset(gap,0,sizeof(gap)); memset(dep,0,sizeof(dep)); memcpy(cur,head,sizeof(head)); int u=start; pre[u]=-1; gap[0]=N; int ans=0; while (dep[start]<N) { if (u==end) { int Min=INF; for (int i=pre[u];i!=-1;i=pre[edge[i^1].to]) if (Min>edge[i].cap-edge[i].flow) Min=edge[i].cap-edge[i].flow; for (int i=pre[u];i!=-1;i=pre[edge[i^1].to]) { edge[i].flow+=Min; edge[i^1].flow-=Min; } u=start; ans+=Min; continue; } bool flag=false; int v; for (int i=cur[u];i!=-1;i=edge[i].next) { v=edge[i].to; if (edge[i].cap-edge[i].flow && dep[v]+1==dep[u]) { flag=true; cur[u]=pre[v]=i; break; } } if (flag) { u=v; continue; } int Min=N; for (int i=head[u];i!=-1;i=edge[i].next) if (edge[i].cap-edge[i].flow && dep[edge[i].to]<Min) { Min=dep[edge[i].to]; cur[u]=i; } gap[dep[u]]--; if (!gap[dep[u]]) return ans; dep[u]=Min+1; gap[dep[u]]++; if (u!=start) u=edge[pre[u]^1].to; } return ans; } int main() { #ifndef ONLINE_JUDGE freopen("C:/Users/asus1/Desktop/IN.txt","r",stdin); #endif int i,j,cas=0; while (~sff(n,m)) { for (i=0;i<n;i++) for (j=0;j<m;j++) sf(mp[i][j]); init(n*m+2); int s=n*m,t=s+1; for (i=0;i<n;i++) { for (j=0;j<m;j++) { if (i>0) addedge(i*m+j,(i-1)*m+j,1); if (j>0) addedge(i*m+j,i*m+j-1,1); if (i<(n-1)) addedge(i*m+j,(i+1)*m+j,1); if (j<(m-1)) addedge(i*m+j,i*m+j+1,1); if (mp[i][j]==1) addedge(i*m+j,t,INF); if (mp[i][j]==2) addedge(s,i*m+j,INF); } } pf("Case %d:\n",++cas); pf("%d\n",sap(s,t,N)); } return 0; }
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Pleasant sheep and big big wolf (hdu 3046 最小割)
标签:pleasant sheep and b hdu 3046 最小割
原文地址:http://blog.csdn.net/u014422052/article/details/46848179