bzoj 1834

建图，费用为零，跑一遍，输出最大流。
再残余图上再建图，0到1建一条容量为K费用为0的边，找出以前的边，从始点到终点再建一条容量无限大，费用为原边费用的边，从0到N跑一遍，输出最小费用。
套的kuangbin的板子：

#include 
using namespace std;
const int MAXN = 1005;
const int MAXM = 100000;
const int INF = 0x3f3f3f3f;
int N;
struct Edge
{int to,next,cap,flow,cost,from;
} edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];void addedge(int u,int v,int cap,int cost)
{edge[tol].from=u;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].from=v;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)
{queueq;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;//printf("i=%d\n",i);//printf("dis[v]=%d dis[u]=%d edge[i].cost=%d cap=%d flow=%d\n",dis[v],dis[u],edge[i].cost,edge[i].cap,edge[i].flow);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;elsereturn true;
}
//返回的是最大流，cost 存的是最小费用
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;
}
//上面是板子，下面是自己写的
int main()
{int M,K,u,v,c,w,co[MAXM];scanf("%d%d%d",&N,&M,&K);memset(head,-1,sizeof(head));while(M--){scanf("%d%d%d%d",&u,&v,&c,&w);addedge(u,v,c,0);co[tol-2]=w;}int ans2;int ans1=minCostMaxflow(1,N,ans2);cout<

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