Codeforces Round #284 (Div. 1) C. Array and Operations(网络流‘最大流)

题目：

http://codeforces.com/problemset/problem/498/C

题意：

给出n个数的数值，m对数（ai，aj），i与j的奇偶性不同，每次选中一对数，是否存在一个数v使得ai与aj都整除，求问最多操作多少次。

思路：

每次选择两数的质因子，则可以保证操作次数最多。

处理出每个数的质因子，接着奇数位的和s连边，偶数位与t连边，有关系的数对连边，跑最大流。

AC.

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <cmath>using namespace std;const int inf = 0x3f3f3f3f;const int maxc = 105;const int maxn = 1010;int n, m;int tot, head[maxn];struct Edge{    int to, next, cap, flow;}edge[maxn*4];void addedge(int u, int v, int w){    edge[tot].to = v; edge[tot].cap = w; edge[tot].flow = 0;    edge[tot].next = head[u]; head[u] = tot++;    edge[tot].to = u; edge[tot].cap = 0; edge[tot].flow = 0;    edge[tot].next = head[v]; head[v] = tot++;}int Q[maxn];int gap[maxn], dep[maxn], cur[maxn];void bfs(int s, int t){    memset(dep, -1, sizeof(dep));    memset(gap, 0, sizeof(gap));    gap[0] = 1;    int fron = 0, rea = 0;    dep[t] = 0;    Q[rea++] = t;    while(fron != rea) {        int u = Q[fron++];        for(int i = head[u]; ~i; i = edge[i].next) {            int v = edge[i].to;            if(dep[v] != -1) continue;            Q[rea++] = v;            dep[v] = dep[u] + 1;            gap[dep[v]]++;        }    }}int S[maxn];int sap(int s, int t, int N){    bfs(s, t);    memcpy(cur, head, sizeof(head));    int top = 0;    int u = s;    int ans = 0;    while(dep[s] < N) {        if(u == t) {            int Min = inf;            int inser;            for(int i = 0; i < top; ++i) {                if(Min > edge[S[i]].cap - edge[S[i]].flow) {                    Min = edge[S[i]].cap - edge[S[i]].flow;                    inser = i;                }            }            for(int i = 0;i < top; ++i) {                edge[S[i]].flow += Min;                edge[S[i]^1].flow -= Min;            }            ans += Min;            top = inser;            u = edge[S[top]^1].to;            continue;        }        bool flag = 0;        int v;        for(int i = cur[u]; ~i; i = edge[i].next) {            v = edge[i].to;            if(edge[i].cap - edge[i].flow && dep[v] + 1 == dep[u]) {                flag = true;                cur[u] = i;                break;            }        }        if(flag) {            S[top++] = cur[u];            u = v;            continue;        }        int Min = N;        for(int i = head[u]; ~i; i = edge[i].next) {            if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min) {                Min = dep[edge[i].to];                cur[u] = i;            }        }        gap[dep[u]]--;        if(!gap[dep[u]]) return ans;        dep[u] = Min + 1;        gap[dep[u]]++;        if(u != s) u = edge[S[--top]^1].to;    }    return ans;}vector<pair<int, int> > v[maxc];int num[maxc][maxn];void deal(int id, int x){    for(int i = 2; i <= sqrt(x)+1; ++i) {        if(x % i == 0) {            while(x % i == 0) {                int ns = v[id].size();                if(!ns) {                    v[id].push_back(make_pair(i, 1));                }                else {                    if(v[id][ns-1].first == i) v[id][ns-1].second++;                    else v[id].push_back(make_pair(i, 1));                }                x /= i;            }        }    }    if(x != 1) {        v[id].push_back(make_pair(x, 1));    }//    printf("id = %d\n", id);//    for(int i = 0; i < v[id].size(); ++i) {//        printf("%d %d\n", v[id][i].first, v[id][i].second);//    }}void init(){    tot = 0;    memset(head, -1, sizeof(head));    for(int i = 0; i <= n; ++i) v[i].clear();}int main(){    //freopen("in", "r", stdin);    while(~scanf("%d %d", &n, &m)) {        init();        int val;        for(int i = 1; i <= n; ++i) {            scanf("%d", &val);            deal(i, val);        }        int cnt = 1;        for(int i = 1; i <= n; ++i) {            for(int j = 0; j < v[i].size(); ++j) {                num[i][j] = cnt++;            }        }        int s = 0, t = cnt;        for(int i = 1; i <= n; ++i) {            for(int j = 0; j < v[i].size(); ++j) {                int c = num[i][j];                if(i % 2) {                    addedge(s, c, v[i][j].second);                }                else {                    addedge(c, t, v[i][j].second);                }            }        }        for(int i = 0; i < m; ++i) {            int x, y;            scanf("%d %d", &x, &y);            if(x % 2 == 0) {                swap(x, y);            }            for(int i = 0; i < v[x].size(); ++i) {                int xx = num[x][i];                for(int j = 0; j < v[y].size(); ++j) {                    int yy = num[y][j];                    if(v[x][i].first == v[y][j].first) {                        addedge(xx, yy, inf);                        //printf("zyz = %d: %d %d\n", v[x][i].first, xx, yy);                    }                }            }        }        int ans = sap(s, t, cnt+1);        printf("%d\n", ans);    }    return 0;}