这道题目不难,二分图匹配建模比较明显。加油吧!相信自己。(自己写的,好开心,40毫秒,比ccz略快)。 尽管算法模版是抄一本书上的,但这次很明显我是背出来的。不算抄。
1 #include<cstdio> 2 #include<iostream> 3 #include<vector> 4 #include<queue> 5 #include<cstring> 6 #define rep(i,j,k) for(int i = j; i <= k; i++) 7 #define maxn 60 8 #define clr(i,j) memset(i,j,sizeof(i)) 9 using namespace std; 10 11 vector<int> g[maxn]; 12 int mx[maxn], my[maxn], dx[maxn], dy[maxn], n1; 13 queue<int> q; 14 bool vis[maxn]; 15 16 bool match(int now) 17 { 18 int s = g[now].size(); 19 rep(i,0,s-1){ 20 int to = g[now][i]; 21 if( !vis[to] && dy[to] == dx[now] + 1 ){ 22 vis[to] = 1; 23 if( !my[to] || match(my[to]) ){ 24 my[to] = now, mx[now] = to; 25 return 1; 26 } 27 } 28 } 29 return 0; 30 } 31 32 int erfen() 33 { 34 clr(mx,0); clr(my,0); int ans = 0; 35 while( 1 ){ 36 clr(dx,0), clr(dy,0); 37 while( !q.empty() ) q.pop(); 38 rep(i,1,n1){ 39 if( !mx[i] ) q.push(i); 40 } 41 bool flag = 1; 42 while( !q.empty() ){ 43 int now = q.front(); q.pop(); 44 int s = g[now].size(); 45 rep(i,0,s-1){ 46 int to = g[now][i]; 47 if( !dy[to] ){ 48 dy[to] = dx[now] + 1; 49 if( my[to] ){ 50 dx[my[to]] = dy[to] + 1; 51 q.push(my[to]); 52 } 53 else flag = 0; 54 } 55 } 56 } 57 if( flag ) break; 58 clr(vis,0); 59 rep(i,1,n1){ 60 if( !mx[i] && match(i) ) ans++; 61 } 62 } 63 return ans; 64 } 65 66 void clear() 67 { 68 rep(i,1,maxn) g[i].clear(); 69 } 70 71 int read() 72 { 73 int s = 0,t = 1; char c = getchar(); 74 while( !isdigit(c) ){ 75 if( c == ‘-‘ ) t = -1; c = getchar(); 76 } 77 while( isdigit(c) ){ 78 s = s * 10 + c- ‘0‘; c = getchar(); 79 } 80 return s * t; 81 } 82 83 bool bed[maxn], home[maxn]; 84 bool used[maxn][maxn]; 85 86 int main() 87 { 88 int t = read(); 89 while( t-- ){ 90 clear(); 91 clr(used,0); clr(bed,0); clr(home,0); 92 int n = read(); 93 n1 = n; 94 rep(i,1,n){ 95 bed[i] = read(); 96 } 97 int tot = 0; 98 rep(i,1,n) { 99 int x = read(); 100 if( bed[i] && x == 1 ) { 101 home[i] = 1; 102 tot++; 103 } 104 } 105 rep(i,1,n){ 106 rep(j,1,n){ 107 int x = read(); 108 if( i == j && bed[i] && !home[i] && !used[i][i] ) { 109 g[i].push_back(i); 110 used[i][i] = 1; 111 } 112 else if( i != j && x ){ 113 if( bed[i] && !home[j] && !used[j][i] ) { 114 g[j].push_back(i); 115 used[j][i] = 1; 116 } 117 if( bed[j] && !home[i] && !used[i][j]) { 118 used[i][j] = 1; 119 g[i].push_back(j); 120 } 121 } 122 } 123 } 124 if( erfen() == n - tot ) printf("%c%c%c\n", 94,95,94); 125 else printf("%c%c%c\n", 84,95,84); 126 } 127 128 return 0; 129 }
原文:http://www.cnblogs.com/83131yyl/p/5106579.html