題意

現(xiàn)在要將5種型號(hào)的衣服分發(fā)給n個(gè)參賽者,然后給出每個(gè)參賽者所需要的衣服的尺碼的大小范圍,在該尺碼范圍內(nèi)的衣服該選手可以接受,再給出這5種型號(hào)衣服各自的數(shù)量,問(wèn)是否存在一種分配方案使得每個(gè)選手都能夠拿到自己尺碼范圍內(nèi)的衣服.

思路

明顯的二分圖多重最大匹配問(wèn)題：每個(gè)點(diǎn)能匹配的邊不再限制1個(gè)，而是多個(gè)。 做法：最大流。雖說(shuō)也有對(duì)應(yīng)的匈牙利算法，但是我還是圖省事用最大流做了。 建圖：源點(diǎn)連接每個(gè)型號(hào)的衣服，容量為能夠匹配的個(gè)數(shù)（數(shù)量），匯點(diǎn)連接每個(gè)隊(duì)員，容量也為能夠匹配的個(gè)數(shù)（1），其他匹配邊容量為1。

代碼

? #include #include #include #include #include #include #include #define MID(x,y) ((x+y)/2) #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; const int MAXV = 35; const int MAXE = 205; const int oo = 0x3fffffff;template struct Dinic{struct flow_node{int u, v;T flow;int opp;int next;}arc[2*MAXE];int vn, en, head[MAXV];int cur[MAXV];int q[MAXV];int path[2*MAXE], top;int dep[MAXV];void init(int n){vn = n;en = 0;mem(head, -1);}void insert_flow(int u, int v, T flow){arc[en].u = u;arc[en].v = v;arc[en].flow = flow;arc[en].next = head[u];head[u] = en ++;arc[en].u = v;arc[en].v = u;arc[en].flow = 0;arc[en].next = head[v];head[v] = en ++;}bool bfs(int s, int t){mem(dep, -1);int lq = 0, rq = 1;dep[s] = 0;q[lq] = s;while(lq < rq){int u = q[lq ++];if (u == t){return true;}for (int i = head[u]; i != -1; i = arc[i].next){int v = arc[i].v;if (dep[v] == -1 && arc[i].flow > 0){dep[v] = dep[u] + 1;q[rq ++] = v;}}}return false;}T solve(int s, int t){T maxflow = 0;while(bfs(s, t)){int i, j;for (i = 1; i <= vn; i ++) cur[i] = head[i];for (i = s, top = 0;;){if (i == t){int mink;T minflow = 0x7fffffff; //要比容量的oo大for (int k = 0; k < top; k ++)if (minflow > arc[path[k]].flow){minflow = arc[path[k]].flow;mink = k;}for (int k = 0; k < top; k ++)arc[path[k]].flow -= minflow, arc[path[k]^1].flow += minflow;maxflow += minflow;top = mink;i = arc[path[top]].u;}for (j = cur[i]; j != -1; cur[i] = j = arc[j].next){int v = arc[j].v;if (arc[j].flow && dep[v] == dep[i] + 1)break;}if (j != -1){path[top ++] = j;i = arc[j].v;}else{if (top == 0) break;dep[i] = -1;i = arc[path[-- top]].u;}}}return maxflow;} }; Dinic dinic;int main(){//freopen("test.in", "r", stdin);//freopen("test.out", "w", stdout);string s;while(cin >> s){if (s == "ENDOFINPUT")break;int n;cin >> n;dinic.init(n+7);for (int i = 1; i <= n; i ++){string tmp;cin >> tmp;dinic.insert_flow(i+5, n+7, 1);for (int j = 0; j < (int)tmp.size(); j ++){switch (tmp[j]){case 'S':dinic.insert_flow(1, i+5, 1);case 'M':dinic.insert_flow(2, i+5, 1);case 'L':dinic.insert_flow(3, i+5, 1);case 'X':dinic.insert_flow(4, i+5, 1);case 'T':dinic.insert_flow(5, i+5, 1);break;}}}for (int i = 1; i <= 5; i ++){int num;cin >> num;dinic.insert_flow(n+6, i, num);}if (dinic.solve(n+6, n+7) == n){puts("T-shirts rock!");}else{puts("I'd rather not wear a shirt anyway...");}cin >> s;}return 0; }

轉(zhuǎn)載于:https://www.cnblogs.com/AbandonZHANG/p/4114074.html

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