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标题效果:鉴于加权值矩阵,带走一个地方的权利值之后,与其相邻的格儿童权利值变0。问多少可以取出到右值。
思维:Amber论文题目。不难建设,图着色。颜色从S连边,还有一种颜色向T连边。再把相邻的格子连边。之后跑最小割,用总权值减去最大流就是答案。
CODE:
#include <queue> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define MAX 400 #define MAXP 40000 #define MAXE 500010 #define INF 0x3f3f3f3f #define S 0 #define T 39999 using namespace std; const int dx[] = {0,1,0,-1,0}; const int dy[] = {0,0,1,0,-1}; int m,n,cnt; int src[MAX][MAX],num[MAX][MAX]; int head[MAXP],total = 1; int next[MAXE],aim[MAXE],flow[MAXE]; int deep[MAXP]; inline void Add(int x,int y,int f) { next[++total] = head[x]; aim[total] = y; flow[total] = f; head[x] = total; } bool BFS() { static queue<int> q; while(!q.empty()) q.pop(); memset(deep,0,sizeof(deep)); deep[S] = 1; q.push(S); while(!q.empty()) { int x = q.front(); q.pop(); for(int i = head[x]; i; i = next[i]) if(flow[i] && !deep[aim[i]]) { deep[aim[i]] = deep[x] + 1; q.push(aim[i]); if(aim[i] == T) return true; } } return false; } int Dinic(int x,int f) { if(x == T) return f; int temp = f; for(int i = head[x]; i; i = next[i]) if(flow[i] && deep[aim[i]] == deep[x] + 1 && temp) { int away = Dinic(aim[i],min(temp,flow[i])); if(!away) deep[aim[i]] = 0; flow[i] -= away; flow[i^1] += away; temp -= away; } return f - temp; } int main() { cin >> m >> n; int sum = 0; for(int i = 1; i <= m; ++i) for(int j = 1; j <= n; ++j) { scanf("%d",&src[i][j]); sum += src[i][j]; num[i][j] = ++cnt; if((i + j)&1) Add(S,num[i][j],src[i][j]),Add(num[i][j],S,0); else Add(num[i][j],T,src[i][j]),Add(T,num[i][j],0); } for(int i = 1; i <= m; ++i) for(int j = 1; j <= n; ++j) { if(((i + j)&1) == 0) continue; for(int k = 1; k <= 4; ++k) { int fx = i + dx[k]; int fy = j + dy[k]; if(!fx || !fy || fx > m || fy > n) continue; Add(num[i][j],num[fx][fy],INF); Add(num[fx][fy],num[i][j],0); } } int max_flow = 0; while(BFS()) max_flow += Dinic(S,INF); cout << sum - max_flow << endl; return 0; }
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原文地址:http://www.cnblogs.com/hrhguanli/p/4656664.html