UVa 11248 - Frequency Hopping 最大流

题目链接：UVa 11248 - Frequency Hopping

题目大意：给你一个有向图，n个点，m条边，1为源点，n为汇点，另给一需求C，问是否存在流的大小为C。如果存在，输出possible，如果不存在，是否可以通过只修改一条边使得条件成立？如果有解，按弧头为第一关键字，弧尾为第二关键字升序输出；无解则输出not possible。

题目分析：先跑一次最大流，如果>=C则直接possible。否则，对所有关键边（容量 = 0（增大流量等于减少容量））容量依次 + C再跑一次最大流，如果条件得到满足则该边为解之一；每一次跑完最大流之后需要对流过的边回溯，这个通过辅助数组记录需要回溯的边就好。关键边的记录也通过一个辅助数组记录。

这次需要对ISAP算法进行略微的修改了，在改变边的容量的时候如果是第一次跑则记录关键边，如果不是则记录需要回溯的边。

PS：总是细节方面问题多多。。。。最后因为忘记对结果排序蛙了一发  +  皿  +。

#include <stdio.h>#include <string.h>#include <algorithm>#define REP(I, X) for(int I = 0; I < X; ++I)#define FF(I, A, B) for(int I = A; I <= B; ++I)#define clear(A, B) memset(A, B, sizeof A)#define copy(A, B) memcpy(A, B, sizeof A)#define min(A, B) ((A) < (B) ? (A) : (B))#define max(A, B) ((A) > (B) ? (A) : (B))using namespace std;typedef long long ll;typedef long long LL;const int oo = 2147483647;const int maxE = 1000000;const int maxN = 105;const int maxQ = 10000;struct Edge{    int v, c, n;};struct Node{    int one, two;};Edge edge[maxE];Node ans[maxE], change[maxE];int adj[maxN], cntE, anscnt, changecnt;int Q[maxE], head, tail, inq[maxN];int d[maxN], num[maxN], cur[maxN], pre[maxN];int s, t, nv;int n, m, need;int get[maxE], getcnt;void addedge(int u, int v, int c){    edge[cntE].v = v; edge[cntE].c = c; edge[cntE].n = adj[u]; adj[u] = cntE++;    edge[cntE].v = u; edge[cntE].c = 0; edge[cntE].n = adj[v]; adj[v] = cntE++;}Node ANS(int i){    Node tmp = {edge[i ^ 1].v, edge[i].v};    return tmp;}Node CHANGE(int i, int c){    Node tmp = {i, c};    return tmp;}int cmp(const Node &a, const Node &b){    return a.one != b.one ? a.one < b.one : a.two < b.two;}void BACK(){    for(int i = 0; i < changecnt; ++i){        int ii = change[i].one, c = change[i].two;        edge[ii].c += c;        edge[ii ^ 1].c -= c;    }}void REV_BFS(){    clear(d, -1);    clear(num, 0);    head = tail = 0;    d[t] = 0;    num[0] = 1;    Q[tail++] = t;    while(head != tail){        int u = Q[head++];        for(int i = adj[u]; ~i; i = edge[i].n){            int v = edge[i].v;            if(~d[v]) continue;            d[v] = d[u] + 1;            num[d[v]]++;            Q[tail++] = v;        }    }}int ISAP(int ch){    copy(cur, adj);    REV_BFS();    int flow = 0, u = pre[s] = s, i;    while(d[s] < nv){        if(u == t){            int f = oo, neck;            for(i = s; i != t; i = edge[cur[i]].v){                if(f > edge[cur[i]].c){                    f = edge[cur[i]].c;                    neck = i;                }            }            for(i = s; i != t; i = edge[cur[i]].v){                edge[cur[i]].c -= f;                edge[cur[i] ^ 1].c += f;                if(ch && !edge[cur[i]].c) get[getcnt++] = cur[i];                else change[changecnt++] = CHANGE(cur[i], f);            }            flow += f;            u = neck;        }        for(i = cur[u]; ~i; i = edge[i].n) if(edge[i].c && d[u] == d[edge[i].v] + 1) break;        if(~i){            cur[u] = i;            pre[edge[i].v] = u;            u = edge[i].v;        }        else{            if(0 == (--num[d[u]])) break;            int mind = nv;            for(i = adj[u]; ~i; i = edge[i].n){                if(edge[i].c && mind > d[edge[i].v]){                    mind = d[edge[i].v];                    cur[u] = i;                }            }            d[u] = mind + 1;            num[d[u]]++;            u = pre[u];        }    }    return flow;}void work(){    int u, v, c, flow;    clear(adj, -1);    cntE = getcnt = anscnt = 0;    s = 1; t = n; nv = t + 1;    while(m--){        scanf("%d%d%d", &u, &v, &c);        addedge(u, v, c);    }    flow = ISAP(1);    if(flow >= need) printf("possible\n");    else{        for(int i = 0; i < getcnt; ++i){            changecnt = 0;            edge[get[i]].c += need;            if(flow + ISAP(0) >= need) ans[anscnt++] = ANS(get[i]);            BACK();            edge[get[i]].c = 0;        }        if(!anscnt) printf("not possible\n");        else{            sort(ans, ans + anscnt, cmp);            printf("possible option:");            for(int i = 0; i < anscnt; ++i){                int u = ans[i].one, v = ans[i].two;                if(i) printf(",");                printf("(%d,%d)", u, v);            }            printf("\n");        }    }}int main(){    int cas = 0;    while(~scanf("%d%d%d", &n, &m, &need) && (n || m || need)){        printf("Case %d: ", ++cas);        work();    }    return 0;}