题目大意

给出一个无向图，一个人会从1号点沿着最短路走到n号点（可能有多条路径），而你需要在某些边的一端设置障碍，使他最终不能够到达n号点，求最小代价，并判断方案是否唯一。

分析

首先建出最短路图。我们可以把原来的边\((u,v)\)拆成两段：\((u,x)\)和\((x,v)\)；割掉\((u,x)\)的代价为\(a[u]\)，割掉\((x,v)\)的代价为\(a[v]\)。我们的目的是用最小代价割掉最短路图上某些边，使得\(1\)走不到\(n\)，这是个最小割模型，于是做一遍最大流就求出了第二问。

问题难在第一问，如何判断最小割是否唯一？

一个不是很显然的结论是：设残量网络中源点\(S\)通过未满流边能够到达的点集为\(S_0\)，通过未满流边能够到达汇点\(T\)的点集为\(T_0\)，若一条边\((u,v)\)满足\(u\in S_0,v\in T_0\)，这条边必定存在于任意一个最小割中。

反证法证明：假设\((u,v)\)不能割，那可以将它的容量视为无穷大，根据题设，我们又能增加\(S\)到\(T\)的流量，这与已经求得最大流矛盾，故\((u,v)\)是必须要割的边。

有了这个结论，我们直接将必割边的代价加起来，如果等于所求最小割，那么方案唯一，否则方案不唯一。

Code

#include <cstdio>
#include <cstring>

typedef long long ll;
const int N = 100007, M = 100007;
const ll INF = (1ll << 60);
ll min(ll a, ll b) { return a < b ? a : b; }

int test;
int n, m, S, T, cnt, s1[N], s2[N];
ll ans, sum, dis[N], a[N], len[M];
int tot, st[N], to[M], nx[M];
int total, head[N], go[M], next[M];
int h, t, q[N], vis[N];
void clear1() { tot = 0; memset(st, 0, sizeof(st)); memset(to, 0, sizeof(to)); memset(nx, 0, sizeof(nx)); }
void clear2() { total = 0; memset(head, 0, sizeof(head)); memset(go, 0, sizeof(go)); memset(next, 0, sizeof(next)); }
void add(int u, int v, ll w) { to[++tot] = v, nx[tot] = st[u], len[tot] = w, st[u] = tot; }
void link(int u, int v) { go[++total] = v, next[total] = head[u], head[u] = total; }

void spfa()
{
    memset(dis, 0x3f, sizeof(dis));
    memset(vis, 0, sizeof(vis));
    h = 1, q[t = 1] = 1, vis[1] = 1, dis[1] = 0;
    while (h <= t)
    {
        int u = q[h++];
        vis[u] = 0;
        for (int i = st[u]; i; i = nx[i])
            if (dis[u] + len[i] < dis[to[i]])
            {
                dis[to[i]] = dis[u] + len[i];
                if (!vis[to[i]]) q[++t] = to[i], vis[to[i]] = 1;
            }
    }
}
void dfs(int u)
{
    vis[u] = 1;
    for (int i = st[u]; i; i = nx[i])
        if (dis[to[i]] + len[i] == dis[u])
        {
            link(u, to[i]);
            if (!vis[to[i]]) dfs(to[i]);
        }
}
int bfs()
{
    memset(dis, 0, sizeof(dis));
    h = 1, q[t = 1] = S, dis[S] = 1;
    while (h <= t)
    {
        int u = q[h++];
        for (int i = st[u]; i; i = nx[i])
            if (!dis[to[i]] && len[i] > 0)
            {
                dis[to[i]] = dis[u] + 1, q[++t] = to[i];
                if (to[i] == T) return 1;
            }
    }
    return 0;
}
ll dinic(int u, ll flow)
{
    if (u == T) return flow;
    ll rest = flow, tmp;
    for (int i = st[u]; i; i = nx[i])
        if (dis[to[i]] == dis[u] + 1 && len[i] > 0)
        {
            tmp = dinic(to[i], min(rest, len[i]));
            if (!tmp) dis[to[i]] = 0;
            len[i] -= tmp, len[i ^ 1] += tmp, rest -= tmp;
            if (!rest) return flow;
        }
    return flow - rest;
}
void getset()
{
    memset(s1, 0, sizeof(s1)); memset(s2, 0, sizeof(s2));
    memset(vis, 0, sizeof(vis));
    h = 1, q[t = 1] = S, vis[S] = 1;
    while (h <= t)
    {
        int u = q[h++]; s1[u] = 1;
        for (int i = st[u]; i; i = nx[i]) if (len[i] > 0 && !vis[to[i]]) q[++t] = to[i], vis[to[i]] = 1;
    }
    memset(vis, 0, sizeof(vis));
    h = 1, q[t = 1] = T, vis[T] = 1;
    while (h <= t)
    {
        int u = q[h++]; s2[u] = 1;
        for (int i = st[u]; i; i = nx[i]) if (len[i ^ 1] > 0 && !vis[to[i]]) q[++t] = to[i], vis[to[i]] = 1;
    }
}

int main()
{
    scanf("%d", &test);
    while (test--)
    {
        clear1(), clear2();
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n - 1; i++) scanf("%lld", &a[i]); a[n] = INF;
        for (int i = 1, u, v, w; i <= m; i++) scanf("%d%d%d", &u, &v, &w), add(u, v, w), add(v, u, w);
        spfa();
        memset(vis, 0, sizeof(vis)), dfs(n);
        clear1(); tot = 1, cnt = n;
        for (int i = 1; i <= n; i++) for (int j = head[i]; j; j = next[j]) add(go[j], ++cnt, a[go[j]]), add(cnt, go[j], 0), add(cnt, i, a[i]), add(i, cnt, 0);
        S = 1, T = n, ans = sum = 0;
        while (bfs()) ans += dinic(S, INF);
        getset();
        cnt = n;
        for (int i = 1; i <= n; i++)
            for (int j = head[i]; j; j = next[j])
            {
                ++cnt;
                if (s1[go[j]] && s2[cnt]) sum += a[go[j]];
                if (s1[cnt] && s2[i]) sum += a[i];
            }
        printf(sum == ans ? "Yes %lld\n" : "No %lld\n", ans);
    }
    return 0;
}

标签：tmp,return,ll,rest,Standard,vis,IO,NOI2013,dis
来源： https://www.cnblogs.com/zjlcnblogs/p/11166405.html