洛谷P2469 星际竞速

上下界费用流比较无脑，提供一种更巧妙的费用流，无需上下界。

  1 #include <cstdio>
  2 #include <algorithm>
  3 #include <queue>
  4 #include <cstring>
  5 
  6 const int N = 1610, M = 1000010, INF = 0x3f3f3f3f;
  7 
  8 struct Edge {
  9     int nex, v, c, len;
 10 }edge[M << 1]; int top = 1;
 11 
 12 int e[N], d[N], vis[N], pre[N], flow[N];
 13 std::queue<int> Q;
 14 
 15 inline void add(int x, int y, int z, int w) {
 16     top++;
 17     edge[top].v = y;
 18     edge[top].c = z;
 19     edge[top].len = w;
 20     edge[top].nex = e[x];
 21     e[x] = top;
 22 
 23     top++;
 24     edge[top].v = x;
 25     edge[top].c = 0;
 26     edge[top].len = -w;
 27     edge[top].nex = e[y];
 28     e[y] = top;
 29     return;
 30 }
 31 
 32 inline bool SPFA(int s, int t) {
 33     memset(d, 0x3f, sizeof(d));
 34     d[s] = 0;
 35     flow[s] = INF;
 36     vis[s] = 1;
 37     Q.push(s);
 38     while(!Q.empty()) {
 39         int x = Q.front();
 40         Q.pop();
 41         vis[x] = 0;
 42         for(int i = e[x]; i; i = edge[i].nex) {
 43             int y = edge[i].v;
 44             if(edge[i].c && d[y] > d[x] + edge[i].len) {
 45                 d[y] = d[x] + edge[i].len;
 46                 pre[y] = i;
 47                 flow[y] = std::min(flow[x], edge[i].c);
 48                 if(!vis[y]) {
 49                     vis[y] = 1;
 50                     Q.push(y);
 51                 }
 52             }
 53         }
 54     }
 55     return d[t] < INF;
 56 }
 57 
 58 inline void update(int s, int t) {
 59     int temp = flow[t];
 60     while(t != s) {
 61         int i = pre[t];
 62         edge[i].c -= temp;
 63         edge[i ^ 1].c += temp;
 64         t = edge[i ^ 1].v;
 65     }
 66     return;
 67 }
 68 
 69 inline int solve(int s, int t, int &cost) {
 70     int ans = 0;
 71     cost = 0;
 72     while(SPFA(s, t)) {
 73         ans += flow[t];
 74         cost += flow[t] * d[t];
 75         update(s, t);
 76     }
 77     return ans;
 78 }
 79 
 80 int main() {
 81     int n, m;
 82     scanf("%d%d", &n, &m);
 83     int s = n * 2 + 1;
 84     int t = s + 1, S = s + 3, T = s + 4, O = s + 5;
 85     for(int i = 1, x; i <= n; i++) {
 86         scanf("%d", &x);
 87         add(O, i, 1, x);
 88         add(i + n, O, INF, 0);
 89         // add(i, i + n, 1, 0);
 90         add(S, i + n, 1, 0);
 91         add(i, T, 1, 0);
 92         add(i + n, t, 1, 0);
 93     }
 94     add(s, O, 1, 0);
 95     for(int i = 1, x, y, z; i <= m; i++) {
 96         scanf("%d%d%d", &x, &y, &z);
 97         if(x > y) {
 98             std::swap(x, y);
 99         }
100         add(x + n, y, INF, z);
101     }
102     add(t, s, INF, 0);
103 
104     int ans;
105     solve(S, T, ans);
106     printf("%d", ans);
107     return 0;
108 }