Uva 11248 Frequency Hopping（最大流）

题意：给定一个有向网络，每条边均有一个容量。问是否存在一个从点1到点N，流量为C的流。如果不存在，是否可以恰好修改一条弧的容量，使得存在这样的流。

思路：先跑一次最大流，如果结果 >= C那么结果是可行的，否则，如果存在某条边满足条件，那么那条边一定是割边，再对求完最大流后的割边枚举，看条件是否满足，

再次求最大流的时候需要保存第一次求完最大流后的结果

#include<cstdio>#include<cstring>#include<queue>#include<vector>#include<map>#include<algorithm>const int maxn = 1e2 + 10;const int INF = 2 * 1e9 + 1e2;using namespace std;struct P {    int to, cap, rev;    P() {}    P(int t, int c, int r) : to(t), cap(c), rev(r) {}};struct E {    int from, to;    E() {}    E(int f, int t) : from(f), to(t) {}    bool operator < (E p) const {        if(from != p.from) return from < p.from;        return to < p.to;    }    bool operator == (E p) const {        return to == p.to && from == p.from;    }};int n, m, C, u, v, tt = 1;int have, num, maxf, limit;int nec[maxn][maxn];int lv[maxn], it[maxn];int Cap[maxn][maxn];E ed[maxn * maxn];vector<P> G[maxn];map<int, int> cap[maxn];void init() {    have = num = limit = maxf = 0;    memset(nec, 0, sizeof(nec));    memset(Cap, 0, sizeof(Cap));    for(int i = 0; i < maxn; i++) {        G[i].clear();        cap[i].clear();    }}void add(int f, int t, int c) {    G[f].push_back(P(t, c, G[t].size()));    G[t].push_back(P(f, 0, G[f].size() - 1));}bool bfs(int s, int t) {    memset(lv, -1, sizeof(lv));    queue<int> q; q.push(s);    lv[s] = 0;    while(!q.empty()) {        int u = q.front(); q.pop();        for(int i = 0; i < G[u].size(); i++) {            P &e = G[u][i];            if(lv[e.to] < 0 && e.cap > 0) {                lv[e.to] = lv[u] + 1;                q.push(e.to);            }        }    }    return lv[t] != -1;}int dfs(int v, int t, int f) {    if(v == t) return f;    for(int &i = it[v]; i < G[v].size(); i++) {        P &e = G[v][i];        if(lv[e.to] <= lv[v] || !e.cap) continue;        int c = dfs(e.to, t, min(e.cap, f));        if(!c) continue;        e.cap -= c;        G[e.to][e.rev].cap += c;        return c;    }    return 0;}int maxflow(int s, int t) {    int f = 0;    while(1) {        if(!bfs(s, t)) return f;        memset(it, 0, sizeof(it));        int fl;        while((fl = dfs(s, t, INF)) > 0)  {            f += fl;            if(maxf + f >= C) return f;        }    }}int main() {    while(scanf("%d %d %d", &n, &m, &C) && n) {        init();        for(int i = 0; i < m; i++) {            int c;            scanf("%d %d %d", &u, &v, &c);            nec[u][v] = 1;            Cap[u][v] += c;        }        for(int i = 1; i <= n; i++) {            for(int j = 1; j <= n; j++) {                if(!Cap[i][j] || i == j) continue;                add(i, j, Cap[i][j]);            }        }        int maxf = maxflow(1, n);        if(maxf >= C) have = 1;        printf("Case %d: ", tt++);        if(have) { printf("possible\n"); continue; }        for(int i = 1; i <= n; i++)            for(int j = 0; j < G[i].size(); j++)                cap[i][j] = G[i][j].cap;        for(int i = 1; i <= n; i++) {            for(int j = 0; j < G[i].size(); j++) {                P &e = G[i][j];                if(e.cap || !nec[i][e.to]) continue;                e.cap = INF;                int res = maxflow(1, n);                for(int ii = 1; ii <= n; ii++) {                    for(int jj = 0; jj < G[ii].size(); jj++) {                        P &st = G[ii][jj];                        st.cap = cap[ii][jj];                    }                }                if(res + maxf < C) continue;                have = 1;                ed[num++] = E(i, e.to);            }        }        if(!have) printf("not possible\n");        else {            sort(ed, ed + num);            num = unique(ed, ed + num) - ed;            printf("possible option:");            for(int i = 0; i < num; i++) {                printf("(%d,%d)", ed[i].from, ed[i].to);                if(i < num - 1) printf(",");                else printf("\n");            }        }    }    return 0;}