[BZOJ3931][CQOI2015]网络吞吐量（SPFA+网络最大流）

首先，跑一遍SPFA，设dis[i]为1到i的最短路径。然后标记出在最短路上的边。判断一条边<u,v>是否在最短路上即判断dis[u]+val<u,v>是否等于dis[v]（val<u,v>表示<u,v>的边权）。然后就可以想到在只由被标记出的边构成的图上跑最大流。
而现在的问题是要限制一个点流出的流量。考虑到这一点，就把一个点拆成一个入点和一个出点，点1的入点为源点，点n的出点为汇点。然后对于每个入点，向它对应的出点连一条边，容量为这个点的吞吐量（流量限制），然后如果存在边<u,v>（被标记），那么就由u的出点向v的入点连一条容量为∞的边。实际上，要限制一个点流出的流量，都可以考虑采用这样的套路。
注意long long。
代码：

#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;inline int read() {    int res = 0; bool bo = 0; char c;    while (((c = getchar()) < '0' || c > '9') && c != '-');    if (c == '-') bo = 1; else res = c - 48;    while ((c = getchar()) >= '0' && c <= '9')        res = (res << 3) + (res << 1) + (c - 48);    return bo ? ~res + 1 : res;}typedef long long ll;const int N = 1e4 + 5, M = 7e5 + 5; const ll INF = 1e18;int n, m, ecnt = 1, nxt[M], adj[N], go[M], ecnt2, nxt2[M], adj2[N], go2[M],val2[M], que[M], lev[N], len, S, T; ll C[N], dis[N], cap[M];bool vis[N];void add_edge(int u, int v, ll w) {    nxt[++ecnt] = adj[u]; adj[u] = ecnt; go[ecnt] = v; cap[ecnt] = w;    nxt[++ecnt] = adj[v]; adj[v] = ecnt; go[ecnt] = u; cap[ecnt] = 0;}void add_edge2(int u, int v, int w) {    nxt2[++ecnt2] = adj2[u]; adj2[u] = ecnt2; go2[ecnt2] = v; val2[ecnt2] = w;    nxt2[++ecnt2] = adj2[v]; adj2[v] = ecnt2; go2[ecnt2] = u; val2[ecnt2] = w;}void SPFA(int st) {    int i; for (i = 1; i <= n; i++) dis[i] = INF;    dis[que[len = 1] = st] = 0;    for (i = 1; i <= len; i++) {        int u = que[i]; vis[u] = 0;        for (int e = adj2[u], v; e; e = nxt2[e])            if (dis[u] + 1ll * val2[e] < dis[v = go2[e]]) {                dis[v] = dis[u] + 1ll * val2[e];                if (!vis[v]) vis[que[++len] = v] = 1;            }    }}void INIT() {    int i; S = 1; T = n << 1; SPFA(1);    for (i = 1; i < T; i += 2) add_edge(i, i + 1, C[(i >> 1) + 1]);    for (i = 1; i <= n; i++) for (int e = adj2[i]; e; e = nxt2[e])        if (dis[i] + 1ll * val2[e] == dis[go2[e]])            add_edge(i << 1, (go2[e] << 1) - 1, INF);}bool bfs() {    int i; memset(lev, -1, sizeof(lev));    lev[que[len = 1] = S] = 0;    for (i = 1; i <= len; i++) {        int u = que[i];        for (int e = adj[u], v; e; e = nxt[e])            if (cap[e] > 0 && lev[v = go[e]] == -1) {                lev[que[++len] = v] = lev[u] + 1;                if (v == T) return 1;            }    }    return 0;}ll dinic(int u, ll flow) {    if (u == T) return flow;    ll res = 0, delta;    for (int e = adj[u], v; e; e = nxt[e])        if (cap[e] > 0 && lev[u] < lev[v = go[e]]) {            delta = dinic(v, min(cap[e], flow - res));            if (delta) {                cap[e] -= delta; cap[e ^ 1] += delta;                res += delta; if (res == flow) break;            }        }    if (res != flow) lev[u] = -1;    return res;}ll solve() {    ll res = 0;    while (bfs()) res += dinic(S, INF);    return res;}int main() {    int i, x, y, z; n = read(); m = read();    for (i = 1; i <= m; i++) x = read(), y = read(),        z = read(), add_edge2(x, y, z);    for (i = 1; i <= n; i++) C[i] = read(); C[1] = C[n] = INF;    printf("%lld\n", (INIT(), solve()));    return 0;}