题目链接
用最多经过(k)条经过(0)的路径覆盖所有点。
定义(ds[i][j])表示从(i)到(j)不经过大于(max(i,j))的点的最短路，显然可以用弗洛伊德求。
然后每个点拆成入出点，连边
源点向(0)的入点连流量k费用0的边，表示最多经过(0)K次
源点向其余每个点的入点连流量1费用0的边
每个(i)的入点向(j(j>i))连流量1费用(ds[i][j])的边
每个点出点向汇点连流量1费用0的边
最小费用即为所求。
#include <cstdio>
#include <queue>
#include <cstring>
#define INF 2147483647
using namespace std;
const int MAXN = 350;
const int MAXM = 40010;
struct Edge{
int from, next, to, rest, cost;
}e[MAXM];
int head[MAXN], num = 1, n, m, k;
inline void Add(int from, int to, int flow, int cost){
e[++num] = (Edge){from, head[from], to, flow, cost}; head[from] = num;
e[++num] = (Edge){to, head[to], from, 0, -cost}; head[to] = num;
}
int s, t, a, b, c, d[MAXM], now, maxflow, mincost;
queue <int> q;
int v[MAXN], dis[MAXN], pre[MAXN], flow[MAXN], ds[MAXN][MAXN];
int re(){
q.push(s);
memset(dis, 127, sizeof dis);
memset(flow, 0, sizeof flow);
dis[s] = 0; pre[t] = 0; flow[s] = INF;
while(q.size()){
now = q.front(); q.pop(); v[now] = 0;
for(int i = head[now]; i; i = e[i].next)
if(e[i].rest && dis[e[i].to] > dis[now] + e[i].cost){
dis[e[i].to] = dis[now] + e[i].cost;
pre[e[i].to] = i; flow[e[i].to] = min(flow[now], e[i].rest);
if(!v[e[i].to]) v[e[i].to] = 1, q.push(e[i].to);
}
}
return pre[t];
}
int main(){
scanf("%d%d%d", &n, &m, &k); s = 345; t = 346;
memset(ds, 63, sizeof ds);
for(int i = 1; i <= m; ++i){
scanf("%d%d%d", &a, &b, &c);
ds[a][b] = ds[b][a] = min(ds[a][b], c);
}
for(int k = 0; k <= n; ++k)
for(int i = 0; i <= n; ++i)
for(int j = 0; j <= n; ++j)
if(k < max(i, j) && ds[i][j] > ds[i][k] + ds[k][j])
ds[i][j] = ds[i][k] + ds[k][j];
for(int i = 0; i <= n; ++i){
Add(s, i, !i ? k : 1, 0);
Add(i + n + 1, t, 1, 0);
for(int j = i + 1; j <= n; ++j)
Add(i, j + n + 1, 1, ds[i][j]);
}
while(re()){
now = pre[t];
while(now){
e[now].rest -= flow[t];
e[now ^ 1].rest += flow[t];
mincost += e[now].cost * flow[t];
now = pre[e[now].from];
}
}
printf("%d
", mincost);
return 0;
}