[SWUST1753] 分配问题（费用流，最优匹配）

题目链接：https://www.oj.swust.edu.cn/problem/show/1753

由于每一个人只能做一件工作，所以要在源汇点处设置容量为1费用为0，在二分图中间设置容量为inf。

而不是源点到人处设置容量为inf。

其实就是最优匹配问题，费用流，这样建图之后权值正负各跑一遍就行了。

 

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 
  4 typedef long long LL;
  5 typedef struct Node {
  6     int u, v, next;
  7     LL c, w;
  8 }Node;
  9 const int maxn = 440;
 10 const int maxm = 80010;
 11 const LL mod = 0x3f3f3f3fLL;
 12 const LL inf = (1LL<<55);
 13 int tot, head[maxn];
 14 LL dist[maxn];
 15 LL cost, flow;
 16 Node e[maxm];
 17 int pre[maxn];
 18 bool visit[maxn];
 19 queue<int> Q;
 20 int S, T, N;
 21 
 22 void init() {
 23     S = T = N = 0;
 24     memset(head, -1, sizeof(head));
 25     tot = 0;
 26 }
 27 
 28 void adde(int u, int v, LL c, LL w) {
 29     e[tot].u = u; e[tot].v = v; e[tot].c = c; e[tot].w = w; e[tot].next = head[u]; head[u] = tot++;
 30     e[tot].u = v; e[tot].v = u; e[tot].c = 0; e[tot].w = -w; e[tot].next = head[v]; head[v] = tot++;
 31 }
 32 bool spfa(int s, int t, int n) {
 33     int i;
 34     for(i = 0; i <= n; i++) {
 35         dist[i] = inf;
 36         visit[i] = 0;
 37         pre[i] = -1;
 38     }
 39     while(!Q.empty()) Q.pop();
 40     Q.push(s);
 41     visit[s] = true;
 42     dist[s] = 0;
 43     pre[s] = -1;
 44     while(!Q.empty()) {
 45         int u = Q.front();
 46         visit[u] = false;
 47         Q.pop();
 48         for(int j = head[u]; j != -1; j = e[j].next) {
 49             if(e[j].c > 0 && dist[u] + e[j].w < dist[e[j].v]) {
 50                 dist[e[j].v] = dist[u] + e[j].w;
 51                 pre[e[j].v] = j;
 52                 if(!visit[e[j].v]) {
 53                     Q.push(e[j].v);
 54                     visit[e[j].v] = true;
 55                 }
 56             }
 57         }
 58     }
 59     if(dist[t] == inf) return false;
 60     else return true;
 61 }
 62 LL ChangeFlow(int t) {
 63     LL det = mod;
 64     int u = t;
 65     while(~pre[u]) {
 66         u = pre[u];
 67         det = min(det, e[u].c);
 68         u = e[u].u;
 69     }
 70     u = t;
 71     while(~pre[u]) {
 72         u = pre[u];
 73         e[u].c -= det;
 74         e[u ^ 1].c += det;
 75         u = e[u].u;
 76     }
 77     return det;
 78 }
 79 LL MinCostFlow(int s, int t, int n) {
 80     LL mincost, maxflow;
 81     mincost = maxflow = 0;
 82     while(spfa(s, t, n)) {
 83         LL det = ChangeFlow(t);
 84         mincost += det * dist[t];
 85         maxflow += det;
 86     }
 87     cost = mincost;
 88     flow = maxflow;
 89     return mincost;
 90 }
 91 
 92 int n;
 93 int w[maxn][maxn];
 94 
 95 int main() {
 96     // freopen("in", "r", stdin);
 97     while(~scanf("%d", &n)) {
 98         init();
 99         S = 0, T = n * 2 + 1, N = T + 1;
100         for(int i = 1; i <= n; i++) {
101             adde(S, i, 1, 0);
102             adde(i+n, T, 1, 0);
103             for(int j = 1; j <= n; j++) {
104                 scanf("%d", &w[i][j]);
105                 adde(i, j+n, inf, w[i][j]);
106             }
107         }
108         cout << MinCostFlow(S, T, N) << endl;
109         init();
110         S = 0, T = n * 2 + 1, N = T + 1;
111         for(int i = 1; i <= n; i++) {
112             adde(S, i, 1, 0);
113             adde(i+n, T, 1, 0);
114             for(int j = 1; j <= n; j++) {
115                 adde(i, j+n, inf, -w[i][j]);
116             }
117         }
118         cout << -MinCostFlow(S, T, N) << endl;
119     }
120     return 0;
121 }

 

posted @ 2017-05-03 15:51  Kirai  阅读(227)  评论(0)  编辑  收藏  举报