bzoj 1266: [AHOI2006]上学路线route

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思路:找出最短路图, 然后求最小割。

  1 #include<bits/stdc++.h>
  2 #define LL long long
  3  
  4 using namespace std;
  5  
  6 const int N = 507;
  7 const int M = 124750;
  8 const int inf = 0x3f3f3f3f;
  9  
 10 struct Edge {
 11     int u, v, t, c, nx;
 12 } edge[2][M << 1];
 13  
 14 int n, m, S, T, tot[2], head[2][N], d[N], in[N], level[N];
 15  
 16 void add(int x, int u, int v, int t, int c) {
 17     edge[x][tot[x]].u = u;
 18     edge[x][tot[x]].v = v;
 19     edge[x][tot[x]].t = t;
 20     edge[x][tot[x]].c = c;
 21     edge[x][tot[x]].nx = head[x][u];
 22     head[x][u] = tot[x]++;
 23 }
 24  
 25 void spfa() {
 26     queue<int> que;
 27     d[1] = 0; que.push(1); in[1] = 1;
 28     while(!que.empty()) {
 29         int u = que.front(); que.pop();
 30         for(int i = head[0][u]; ~i; i = edge[0][i].nx) {
 31             int v = edge[0][i].v, t = edge[0][i].t;
 32             if(d[u] + t < d[v]) {
 33                 d[v] = d[u] + t;
 34                 if(!in[v]) que.push(v);
 35             }
 36         }
 37         in[u] = 0;
 38     }
 39 }
 40  
 41 bool bfs() {
 42     memset(level, 0, sizeof(level));
 43     queue<int> que;
 44     que.push(S); level[S] = 1;
 45     while(!que.empty()) {
 46         int u = que.front(); que.pop();
 47         if(u == T) return true;
 48         for(int i = head[1][u]; ~i; i = edge[1][i].nx) {
 49             int v = edge[1][i].v;
 50             if(!level[v] && edge[1][i].c > 0) {
 51                 que.push(v);
 52                 level[v] = level[u] + 1;
 53             }
 54         }
 55     }
 56     return false;
 57 }
 58  
 59 int dfs(int u, int p) {
 60     if(u == T) return p;
 61     int ret = 0;
 62     for(int i = head[1][u]; ~i; i = edge[1][i].nx) {
 63         int v = edge[1][i].v, c = edge[1][i].c;
 64         if(c <= 0 || level[v] != level[u] + 1) continue;
 65         int f = dfs(v, min(p - ret, c));
 66         ret += f;
 67         edge[1][i].c -= f;
 68         edge[1][i ^ 1].c += f;
 69         if(ret == p) break;
 70     }
 71     if(!ret) level[u] = 1;
 72     return ret;
 73 }
 74  
 75 int Dinic() {
 76     int ans = 0;
 77     while(bfs()) ans += dfs(S, inf);
 78     return ans;
 79 }
 80 int main() {
 81     memset(head, -1, sizeof(head));
 82     memset(d, inf, sizeof(d));
 83     scanf("%d%d", &n, &m);
 84     S = 1, T = n;
 85     for(int i = 1; i <= m; i++) {
 86         int u, v, t, c;
 87         scanf("%d%d%d%d", &u, &v, &t, &c);
 88         add(0, u, v, t, c);
 89         add(0, v, u, t, c);
 90     }
 91  
 92     spfa();
 93  
 94     for(int i = 0; i < tot[0]; i++) {
 95         int u = edge[0][i].u;
 96         int v = edge[0][i].v;
 97         int t = edge[0][i].t;
 98         int c = edge[0][i].c;
 99  
100         if(d[u] + t == d[v]) {
101             add(1, u, v, t, c);
102             add(1, v, u, t, 0);
103         }
104     }
105  
106     int ans = Dinic();
107  
108     printf("%d\n", d[n]);
109     printf("%d\n", ans);
110     return 0;
111 }

 

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