题意
现在有n个任务，两个机器A和B，每个任务要么在A上完成，要么在B上完成，而且知道每个任务在A和B机器上完成所需要的费用。然后再给m行，每行 a，b，w三个数字。表示如果a任务和b任务不在同一个机器上工作的话，需要额外花费w。现在要求出完成所有任务最小的花费是多少。
思路
上次做的构图题是基于割截断s->t流与题目联系的，而这道题的构图则是基于割把流网络的点划分成了S、T点集，并且不同点集间的边都是割边。这是目前我所接触到的最小割的建模类型。
回到本题构图：源点向任务连一条Ai容量的边，任务向汇点连一条Bi容量的边，如果两任务在不同Core上需要代价则连一条(i,j,cost)的无向边。
代码
#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 = 505; const int MAXE = 20005; const int oo = 0x3fffffff; struct node{ int u, v, flow; int opp; int next; }; struct Dinic{ node arc[MAXE]; int vn, en, head[MAXV]; //vn点个数(包括源点汇点),en边个数 int cur[MAXV]; //当前弧 int q[MAXV]; //bfs建层次图时的队列 int path[MAXE], top; //存dfs当前最短路径的栈 int dep[MAXV]; //各节点层次 void init(int n){ vn = n; en = 0; mem(head, -1); } void insert_flow(int u, int v, int flow){ arc[en].u = u; arc[en].v = v; arc[en].flow = flow; arc[en].opp = en + 1; arc[en].next = head[u]; head[u] = en ++; arc[en].u = v; arc[en].v = u; arc[en].flow = 0; //反向弧 arc[en].opp = en - 1; 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; } int solve(int s, int t){ int 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; int minflow = 0x3fffffff; 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[arc[path[k]].opp].flow += minflow; maxflow += minflow; top = mink; //arc[mink]这条边流量变为0, 则直接回溯到该边的起点即可(这条边将不再包含在增广路内). 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; bool vis[MAXV]; bool reach(int u, int p){ vis[u] = 1; if (u == p) return true; for (int i = dinic.head[u]; i != -1; i = dinic.arc[i].next){ if (i % 2 == 1) continue; int v = dinic.arc[i].v; if (vis[v] || dinic.arc[i].flow <= 0) continue; if (reach(v, p)) return true; } return false; } int work(int n){ vector
seg; for (int i = 0; i < dinic.en; i += 2){ if (dinic.arc[i].flow == 0){ mem(vis, 0); int u = dinic.arc[i].u; int v = dinic.arc[i].v; if (reach(n+1, u) && reach(v, n+2)){ seg.push_back(i/2+1); } } } for (int i = 0; i < (int)seg.size(); i ++){ if (i == 0) printf("%d", seg[i]); else printf(" %d", seg[i]); } puts(""); } int main(){ //freopen("test.in", "r", stdin); //freopen("test.out", "w", stdout); int n, m, l; while(scanf("%d %d %d", &n, &m, &l), n){ dinic.init(n+m+2); for (int i = 0; i < l; i ++){ int u,v,w; scanf("%d %d %d", &u, &v, &w); dinic.insert_flow(u==0?n+m+2:u, v==0?n+m+2:v, w); } for (int i = 1; i <= n; i ++){ dinic.insert_flow(n+m+1, i, oo); } dinic.solve(n+m+1, n+m+2); work(n+m); } return 0; }
