hdu-4292.food(绫籨ining缃戠粶娴佸缓鍥?
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鏍囩锛?a href='http://www.mamicode.com/so/1/inpu' title='inpu'>inpu php rem ESS bool some enc person bsp
Food
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9289 Accepted Submission(s): 3019
Problem Description
銆€銆€You, a part-time dining service worker in your college’s dining hall, are now confused with a new problem: serve as many people as possible.
銆€銆€The issue comes up as people in your college are more and more difficult to serve with meal: They eat only some certain kinds of food and drink, and with requirement unsatisfied, go away directly.
銆€銆€You have prepared F (1 <= F <= 200) kinds of food and D (1 <= D <= 200) kinds of drink. Each kind of food or drink has certain amount, that is, how many people could this food or drink serve. Besides, You know there’re N (1 <= N <= 200) people and you too can tell people’s personal preference for food and drink.
銆€銆€Back to your goal: to serve as many people as possible. So you must decide a plan where some people are served while requirements of the rest of them are unmet. You should notice that, when one’s requirement is unmet, he/she would just go away, refusing any service.
銆€銆€The issue comes up as people in your college are more and more difficult to serve with meal: They eat only some certain kinds of food and drink, and with requirement unsatisfied, go away directly.
銆€銆€You have prepared F (1 <= F <= 200) kinds of food and D (1 <= D <= 200) kinds of drink. Each kind of food or drink has certain amount, that is, how many people could this food or drink serve. Besides, You know there’re N (1 <= N <= 200) people and you too can tell people’s personal preference for food and drink.
銆€銆€Back to your goal: to serve as many people as possible. So you must decide a plan where some people are served while requirements of the rest of them are unmet. You should notice that, when one’s requirement is unmet, he/she would just go away, refusing any service.
Input
銆€銆€There are several test cases.
銆€銆€For each test case, the first line contains three numbers: N,F,D, denoting the number of people, food, and drink.
銆€銆€The second line contains F integers, the ith number of which denotes amount of representative food.
銆€銆€The third line contains D integers, the ith number of which denotes amount of representative drink.
銆€銆€Following is N line, each consisting of a string of length F. 顏塭 jth character in the ith one of these lines denotes whether people i would accept food j. “Y” for yes and “N” for no.
銆€銆€Following is N line, each consisting of a string of length D. 顏塭 jth character in the ith one of these lines denotes whether people i would accept drink j. “Y” for yes and “N” for no.
銆€銆€Please process until EOF (End Of File).
銆€銆€For each test case, the first line contains three numbers: N,F,D, denoting the number of people, food, and drink.
銆€銆€The second line contains F integers, the ith number of which denotes amount of representative food.
銆€銆€The third line contains D integers, the ith number of which denotes amount of representative drink.
銆€銆€Following is N line, each consisting of a string of length F. 顏塭 jth character in the ith one of these lines denotes whether people i would accept food j. “Y” for yes and “N” for no.
銆€銆€Following is N line, each consisting of a string of length D. 顏塭 jth character in the ith one of these lines denotes whether people i would accept drink j. “Y” for yes and “N” for no.
銆€銆€Please process until EOF (End Of File).
Output
銆€銆€For each test case, please print a single line with one integer, the maximum number of people to be satisfied.
Sample Input
4 3 3
1 1 1
1 1 1
YYN
NYY
YNY
YNY
YNY
YYN
YYN
NNY
Sample Output
3
Source
濡傛灉浣犲仛杩噋oj3281浣犲簲璇ユ竻闄や粬浠緢鍍忥紝濡傛灉浣犳病鍋氳繃鍙互閫夋嫨鍏堢湅鐪嬮偅閬撴洿绠€鍗曚竴鐐圭殑棰樼洰銆?/div>
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涓嬮潰杩欐槸鎴戣嚜宸卞啓鐨勭涓€绉嶅仛娉曪紝寤虹珛寰堝澶氫綑鐨勮妭鐐癸紝鍥犱负鎴戞兂鐫€闇€瑕佺粨鐐瑰閲忛檺鍒讹紝鎵€浠ユ瘡涓粨鐐归兘鎷嗕簡锛屽仛瀹屼箣鍚庣湅鍒汉鐨勪唬鐮佸彂鐜板師鏉ュ彲浠ユ湁鏇寸畝鍗曠殑寤哄浘鏂规硶锛岄偅涔堢湅鏈€鍚庡惂銆?/div>
1 /* 2 缁撶偣0 ~ n - 1瀛樺乏鐗涚粨鐐? 3 缁撶偣n ~ 2 * n - 1瀛樺彸鐗涚粨鐐? 4 缁撶偣2 * n ~ 2 * n + f - 1瀛樺乏椋熺墿 5 缁撶偣2 * n + f ~ 2 * n + f * 2 - 1瀛樺彸椋熺墿 6 缁撶偣2 * n + 2 * f ~ 2 * n + 2 * f + d - 1瀛橀ギ鏂欏乏 7 缁撶偣2 * n + 2 * f + d ~ 2 * n + 2 * f + d * 2 - 1瀛橀ギ鏂欏彸 8 缁撶偣2 * n + 2 * f + 2 * d涓簊锛宼 = s = 1銆? 9 */ 10 11 #include <cstdio> 12 #include <cstring> 13 #include <algorithm> 14 using namespace std; 15 16 const int maxn = 200 + 5, maxm = 1000 * 1000 + 5, inf = 0x3f3f3f3f; 17 int numf[maxn], numd[maxn]; 18 char str[maxn]; 19 20 int tot, head[maxn << 3], que[maxn << 3], dep[maxn << 3], cur[maxn << 3], sta[maxn << 3]; 21 22 struct Edge 23 int to, cap, flow, next, from; 24 edge[maxm << 1]; 25 26 void init() 27 tot = 2; 28 memset(head, -1, sizeof head); 29 30 31 void addedge(int u, int v, int w) 32 edge[tot].to = v; edge[tot].cap = w; edge[tot].flow = 0; edge[tot].from = u; 33 edge[tot].next = head[u]; head[u] = tot ++; 34 edge[tot].to = u; edge[tot].cap = 0; edge[tot].flow = 0; edge[tot].from = v; 35 edge[tot].next = head[v]; head[v] = tot ++; 36 37 38 bool bfs(int s, int t, int n) 39 memset(dep, -1, sizeof dep[0] * (n + 1)); 40 int front = 0, tail = 0; 41 dep[s] = 0; 42 que[tail ++] = s; 43 while(front < tail) 44 int u = que[front ++]; 45 for(int i = head[u]; ~i; i = edge[i].next) 46 int v = edge[i].to; 47 if(edge[i].cap > edge[i].flow && dep[v] == -1) 48 dep[v] = dep[u] + 1; 49 if(v == t) return true; 50 que[tail ++] = v; 51 52 53 54 return false; 55 56 57 int dinic(int s, int t, int n) 58 int maxflow = 0; 59 while(bfs(s, t, n)) 60 for(int i = 0; i <= n; i ++) cur[i] = head[i]; 61 int u = s, tail = 0; 62 while(~cur[s]) 63 if(u == t) 64 int tp = inf; 65 for(int i = tail - 1; i >= 0; i --) 66 tp = min(tp, edge[sta[i]].cap - edge[sta[i]].flow); 67 maxflow += tp; 68 for(int i = tail - 1; i >= 0; i --) 69 edge[sta[i]].flow += tp; 70 edge[sta[i] ^ 1].flow -= tp; 71 if(edge[sta[i]].cap - edge[sta[i]].flow == 0) tail = i; 72 73 u = edge[sta[tail] ^ 1].to; 74 else if(~ cur[u] && edge[cur[u]].cap > edge[cur[u]].flow && dep[u] + 1 == dep[edge[cur[u]].to]) 75 sta[tail ++] = cur[u]; 76 u = edge[cur[u]].to; 77 else 78 while(u != s && cur[u] == -1) 79 u = edge[sta[-- tail] ^ 1].to; 80 cur[u] = edge[cur[u]].next; 81 82 83 84 return maxflow; 85 86 87 int main() 88 int n, f, d; 89 while(~scanf("%d %d %d", &n, &f, &d)) 90 init(); 91 int s = 2 * n + 2 * f + d * 2, t = s + 1;//瓒呯骇婧愮偣鍜岃秴绾ф眹鐐?/span> 92 for(int i = 0; i < f; i ++) 93 scanf("%d", &numf[i]); 94 addedge(s, 2 * n + i, inf);//瓒呯骇婧愮偣鍒伴鐗╁乏 95 addedge(2 * n + i, 2 * n + f + i, numf[i]);//椋熺墿宸﹀埌椋熺墿鍙?/span> 96 97 for(int i = 0; i < d; i ++) 98 scanf("%d", &numd[i]); 99 addedge(2 * n + 2 * f + i, 2 * n + 2 * f + d + i, numd[i]);//楗枡宸﹀埌楗枡鍙?/span> 100 addedge(2 * n + 2 * f + d + i, t, inf);//楗枡鍙冲埌瓒呯骇姹囩偣 101 102 for(int i = 0; i < n; i ++) 103 addedge(i, n + i, 1);//宸︾墰鍒板彸鐗?/span> 104 105 for(int i = 0; i < n; i++) 106 scanf("%s", str); 107 for(int j = 0; j < f; j ++) 108 if(str[j] == 鈥?/span>Y鈥?/span>) 109 addedge(2 * n + f + j, i, 1);//浠庨鐗╁彸鍒板乏鐗?/span> 110 111 112 for(int i = 0; i < n; i ++) 113 scanf("%s", str); 114 for(int j = 0; j < d; j ++) 115 if(str[j] == 鈥?/span>Y鈥?/span>) 116 addedge(n + i, 2 * n + 2 * f + j, 1);//浠庡彸鐗涘埌宸﹂ギ鏂?/span> 117 118 119 // for(int i = 2; i <= tot; i ++) 120 // printf("%d -> %d\n", edge[i].from, edge[i].to); 121 // 122 printf("%d\n", dinic(s, t, 2 * n + 2 * f + 2 * d + 2)); 123 124 return 0; 125
涓嬮潰杩欑寤哄浘鏂规硶鍜屼笂闈㈢殑绫讳技锛屽彧鏄笂鍥炬槸鎷嗙偣闄愬埗鐐规祦閲忥紝鑰屾垜浠煡閬撳浜庢瘡涓€浠堕鐗╋紝濡傛灉鎴戜滑鏈変竴涓汉閫夊彇瀹冿紝閭ｄ箞瀹冨繀瀹氭槸鍙€夊彇浜嗕竴浠讹紝鍥犱负鍚庨潰鎷嗙偣n鍐冲畾鐨勶紝閭ｄ箞涔熷氨鏄瘡涓汉鍙兘鍙栦粬鎵€鍠滄椋熺墿涓殑涓€绉嶄腑鐨勪竴涓紝鎵€浠ユ垜浠彧闇€瑕佸鎴戜滑鑳藉鎻愪緵鐨勬煇绉嶉鐗╅檺閲忓氨鍙互浜嗭紝涔熷氨鏄粠婧愮偣鍒版煇绉嶉鐗╂潈鍊间负food_num锛岃繖鏍峰氨鍙互闄愬埗浣忔瘡绉嶉鐗╃殑鐢ㄩ噺浜嗭紝鎺ョ潃鐪嬮ギ鏂欙紝濡傛灉鏌愪釜浜洪€夊彇浜嗕竴涓ギ鏂欙紝閭ｄ箞浠栦篃鍙兘閫夊彇杩欎竴绉嶉ギ鏂欎腑鐨勪竴鐡讹紝鍥犱负鍓嶉潰宸茬粡瀵筺鎷嗙偣瀵艰嚧瀹冭兘鎵╁睍鐨勬祦涔熷彧鏈?锛屾墍浠ュ鑷村ス閫夌殑楗枡涔熸槸1瀵?锛屾墍浠ユ兂瑕侀檺鍒堕ギ鏂欑殑涓暟锛屼篃鍙渶瑕佹棤闄愮储鍙栵紝鐩村埌鏈€鍚庢棤娉曟祦鍒版眹鐐瑰氨ok锛岄偅涔熷氨鏄粠楗枡鍒版眹鐐规潈鍊间负drink_num銆?/p>
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 using namespace std; 5 6 const int maxn = 200 + 5, maxm = 1000 * 1000 + 5, inf = 0x3f3f3f3f; 7 int numf[maxn], numd[maxn]; 8 char str[maxn]; 9 10 int tot, head[maxn << 3], que[maxn << 3], dep[maxn << 3], cur[maxn << 3], sta[maxn << 3]; 11 12 struct Edge 13 int to, cap, flow, next, from; 14 edge[maxm << 1]; 15 16 void init() 17 tot = 2; 18 memset(head, -1, sizeof head); 19 20 21 void addedge(int u, int v, int w) 22 edge[tot].to = v; edge[tot].cap = w; edge[tot].flow = 0; edge[tot].from = u; 23 edge[tot].next = head[u]; head[u] = tot ++; 24 edge[tot].to = u; edge[tot].cap = 0; edge[tot].flow = 0; edge[tot].from = v; 25 edge[tot].next = head[v]; head[v] = tot ++; 26 27 28 bool bfs(int s, int t, int n) 29 memset(dep, -1, sizeof dep[0] * (n + 1)); 30 int front = 0, tail = 0; 31 dep[s] = 0; 32 que[tail ++] = s; 33 while(front < tail) 34 int u = que[front ++]; 35 for(int i = head[u]; ~i; i = edge[i].next) 36 int v = edge[i].to; 37 if(edge[i].cap > edge[i].flow && dep[v] == -1) 38 dep[v] = dep[u] + 1; 39 if(v == t) return true; 40 que[tail ++] = v; 41 42 43 44 return false; 45 46 47 int dinic(int s, int t, int n) 48 int maxflow = 0; 49 while(bfs(s, t, n)) 50 for(int i = 0; i <= n; i ++) cur[i] = head[i]; 51 int u = s, tail = 0; 52 while(~cur[s]) 53 if(u == t) 54 int tp = inf; 55 for(int i = tail - 1; i >= 0; i --) 56 tp = min(tp, edge[sta[i]].cap - edge[sta[i]].flow); 57 maxflow += tp; 58 for(int i = tail - 1; i >= 0; i --) 59 edge[sta[i]].flow += tp; 60 edge[sta[i] ^ 1].flow -= tp; 61 if(edge[sta[i]].cap - edge[sta[i]].flow == 0) tail = i; 62 63 u = edge[sta[tail] ^ 1].to; 64 else if(~ cur[u] && edge[cur[u]].cap > edge[cur[u]].flow && dep[u] + 1 == dep[edge[cur[u]].to]) 65 sta[tail ++] = cur[u]; 66 u = edge[cur[u]].to; 67 else 68 while(u != s && cur[u] == -1) 69 u = edge[sta[-- tail] ^ 1].to; 70 cur[u] = edge[cur[u]].next; 71 72 73 74 return maxflow; 75 76 77 int main() 78 int n, f, d; 79 while(~scanf("%d %d %d", &n, &f, &d)) 80 init(); 81 int s = 2 * n + f + d, t = s + 1;//瓒呯骇婧愮偣鍜岃秴绾ф眹鐐?/span> 82 for(int i = 0; i < f; i ++) 83 scanf("%d", &numf[i]); 84 addedge(s, n * 2 + i, numf[i]); 85 86 for(int i = 0; i < d; i ++) 87 scanf("%d", &numd[i]); 88 addedge(n * 2 + f + i, t, numd[i]); 89 90 for(int i = 0; i < n; i++) 91 addedge(i, n + i, 1);//宸︾墰鍒板彸鐗?/span> 92 scanf("%s", str); 93 for(int j = 0; j < f; j ++) 94 if(str[j] == 鈥?/span>Y鈥?/span>) 95 addedge(2 * n + j, i, 1); 96 97 98 for(int i = 0; i < n; i ++) 99 scanf("%s", str); 100 for(int j = 0; j < d; j ++) 101 if(str[j] == 鈥?/span>Y鈥?/span>) 102 addedge(n + i, 2 * n + f + j, 1);//浠庡彸鐗涘埌楗枡 103 104 105 // for(int i = 2; i <= tot; i ++) 106 // printf("%d -> %d\n", edge[i].from, edge[i].to); 107 // 108 printf("%d\n", dinic(s, t, 2 * n + f + d + 2)); 109 110 return 0; 111
涓嶅緱涓嶆壙璁よ繖绉嶅仛娉曠‘瀹炶妭鐪佷簡寰堝绌洪棿鍛€銆?/p>
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