題干：

Far, far away there is a world known as Selfishland because of the nature of its inhabitants. Hard times have forced the cities of Selfishland to exchange goods among each other. C1 cities are willing to sell some goods and the other C2 cities are willing to buy some goods (each city can either sell or buy goods, but not both). There would be no problem if not for the selfishness of the cities. Each selling city will sell its goods to one city only, and each buying city will buy goods from one city only.?

Your goal is to connect the selfish cities in such a way that the amount of exchanged goods is maximalized.

Input

The first line contains a positive integer t<=1000 indicating the number of test cases. Each test case is an instance of the problem defined above. The first line of each test case is a pair of positive integers C1 and C2 (the number of cities wanting to sell their goods C1<=100 and the number of cities wanting to buy goods C2<=100). The lines that follow contain a sequence of (c1,c2,g) trios ending with three zeros. (c1,c2,g) means that the city c1 can offer the city c2 the amount of g<=100 goods.

Output

For each test case print the maximal amount of goods exchanged.

Example

Input: 3 3 2 1 1 10 2 1 19 2 2 11 3 2 1 0 0 0 4 4 1 1 6 1 2 6 2 1 8 2 3 9 2 4 8 3 2 8 4 3 7 0 0 0 3 2 1 1 10 2 1 21 2 2 11 3 2 1 0 0 0Output: 21 29 22

題目大意：（摘抄）

? ? 有A個城市賣東西，B個城市買東西，一個城市只能賣東西給一個城市，一個城市只能從一個城市買東西，問最多有多少貨物被交易

解題報告：

? ?直接造一個超級源點超級匯點建圖跑MCMF就可以了，，注意邊權取反，，因為要跑最大費用流，答案也取反就行了，，

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long using namespace std;const int MAXN = 70000; const int MAXM = 100005; const int INF = 0x3f3f3f3f; struct Edge {int to,next,cap,flow,cost; } edge[MAXM]; int head[MAXN],tol; int pre[MAXN],dis[MAXN]; bool vis[MAXN]; int N;//節點總個數，節點編號從 1 ～ N void init(int n) {N = n;tol = 0;memset(head, -1,sizeof(head)); } void addedge(int u,int v,int cap,int cost) {edge[tol].to = v;edge[tol].cap = cap;edge[tol].cost = cost;edge[tol].flow = 0;edge[tol].next = head[u];head[u] = tol++;edge[tol].to = u;edge[tol].cap = 0;edge[tol].cost = -cost;edge[tol].flow = 0;edge[tol].next = head[v];head[v] = tol++; } bool spfa(int s,int t) {queue<int>q;for(int i = 0; i <= N; i++) {dis[i] = INF;vis[i] = false;pre[i] = -1;}dis[s] = 0;vis[s] = true;q.push(s);while(!q.empty()) {int u = q.front();q.pop();vis[u] = false;for(int i = head[u]; i !=-1; i = edge[i].next) {int v = edge[i].to;if(edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i].cost ) {dis[v] = dis[u] + edge[i].cost;pre[v] = i;if(!vis[v]) {vis[v] = true;q.push(v);}}}}if(pre[t] ==-1)return false;else return true;} //返回的是最大流，cost 存的是最小費用 int minCostMaxflow(int s,int t,int &cost) {int flow = 0;cost = 0;while(spfa(s,t)) {int Min = INF;for(int i = pre[t]; i !=-1; i = pre[edge[i^1].to]) {if(Min > edge[i].cap-edge[i].flow)Min = edge[i].cap-edge[i].flow;}for(int i = pre[t]; i !=-1; i = pre[edge[i^1].to]) {edge[i].flow += Min;edge[i^1].flow-= Min;cost += edge[i].cost * Min;}flow += Min;}return flow; } int main() {int n1,n2,n;int t;cin>>t;while(t--) {scanf("%d%d",&n1,&n2);//int st=0,ed=n1+n2+1;int st = 0,ed=n1+n2+1;init(ed+1);int a,b,w;while(scanf("%d%d%d",&a,&b,&w)) {if(a+b+w==0) break;b+=n1; // printf("%d %d %d\n",a,b,w);addedge(a,b,1,-w);}for(int i = 1; i<=n1; i++) addedge(st,i,1,0);for(int i = n1+1; i<=n1+n2; i++) addedge(i,ed,1,0);int cost;int ans = minCostMaxflow(st,ed,cost);printf("%d\n",-cost);}return 0 ; }

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總結

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