标签:nbsp lib 整数 printf strong ;; img 规模 name
[codevs1022]覆盖
试题描述
有一个N×M的单位方格中,其中有些方格是水塘,其他方格是陆地。如果要用1×2的矩阵区覆盖(覆盖过程不容许有任何部分重叠)这个陆地,那么最多可以覆盖多少陆地面积。

输入
输入文件的第一行是两个整数N,M (1<=N,M<=100),第二行为一个整数K( K<=50),接下来的K行,每行两个整数X,Y表示K个水塘的行列位置。(1<=X<=N,1<=Y<=M)。
输出
输出所覆盖的最大面积块(1×2面积算一块)。
输入示例
4 4 6 1 1 1 4 2 2 4 1 4 2 4 4
输出示例
4
数据规模及约定
见“输入”
题解
黑白染色后,挖去那几个被排除的点跑二分图匹配。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;
int read() {
int x = 0, f = 1; char c = getchar();
while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }
while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); }
return x * f;
}
#define maxn 10010
#define maxm 160010
#define oo 2147483647
struct Edge {
int from, to, flow;
Edge() {}
Edge(int _1, int _2, int _3): from(_1), to(_2), flow(_3) {}
} ;
struct Dinic {
int n, m, s, t, head[maxn], next[maxm];
Edge es[maxm];
int hd, tl, Q[maxn], vis[maxn];
int cur[maxn];
void init(int nn) {
n = nn; m = 0;
memset(head, -1, sizeof(head));
return ;
}
void AddEdge(int a, int b, int c) {
es[m] = Edge(a, b, c); next[m] = head[a]; head[a] = m++;
return ;
}
bool BFS() {
memset(vis, 0, sizeof(vis)); vis[s] = 1;
hd = tl = 0; Q[++tl] = s;
while(hd < tl) {
int u = Q[++hd];
for(int i = head[u]; i != -1; i = next[i]) {
Edge& e = es[i];
if(!vis[e.to] && e.flow) {
vis[e.to] = vis[u] + 1;
Q[++tl] = e.to;
}
}
}
return vis[t] > 1;
}
int DFS(int u, int a) {
if(u == t || !a) return a;
int flow = 0, f;
for(int& i = cur[u]; i != -1; i = next[i]) {
Edge& e = es[i];
if(vis[e.to] == vis[u] + 1 && (f = DFS(e.to, min(a, e.flow)))) {
flow += f; a -= f;
e.flow -= f; es[i^1].flow += f;
if(!a) return flow;
}
}
return flow;
}
int MaxFlow(int ss, int tt) {
s = ss; t = tt;
int flow = 0;
while(BFS()) {
for(int i = 1; i <= n; i++) cur[i] = head[i];
flow += DFS(s, oo);
}
return flow;
}
} sol;
#define maxs 110
bool Map[maxs][maxs];
int main() {
int n = read(), m = read(), k = read();
sol.init(n * m + 2); int s = n * m + 1, t = s + 1;
for(int i = 1; i <= k; i++) {
int a = read() - 1, b = read() - 1;
Map[a][b] = 1;
}
for(int i = 0; i < n; i++)
for(int j = 0; j < m - 1; j++) if(!Map[i][j] && !Map[i][j+1]) {
int id = i * m + j + 1, rid = i * m + j + 2;
sol.AddEdge(id, rid, 1); sol.AddEdge(rid, id, 1);
}
for(int i = 0; i < n - 1; i++)
for(int j = 0; j < m; j++) if(!Map[i][j] && !Map[i+1][j]) {
int id = i * m + j + 1, did = (i+1) * m + j + 1;
sol.AddEdge(id, did, 1); sol.AddEdge(did, id, 1);
}
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++) if(!Map[i][j]) {
int id = i * m + j + 1;
if((i & 1) ^ (j & 1)) sol.AddEdge(s, id, 1), sol.AddEdge(id, s, 0);
else sol.AddEdge(id, t, 1), sol.AddEdge(t, id, 0);
}
printf("%d\n", sol.MaxFlow(s, t));
return 0;
}
标签:nbsp lib 整数 printf strong ;; img 规模 name
原文地址:http://www.cnblogs.com/xiao-ju-ruo-xjr/p/6034873.html