UVa 563

题目分析

因为每个点有限制，很容易想到拆点，然后求最大流判断是否等于b即可。现在已经很容易手打ISAP网络流算法了。。

#include <queue>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 1e5+100;const int maxm = 1e6+100;const int INF = 0x3f3f3f3f;struct Edge{    int to, cap, next;}e[maxm<<1];int head[maxn], tot;int num[maxn], d[maxn], pre[maxn], cur[maxn];int sink, source, nv;int s, a, b;void addedge(int from, int to){    e[tot].to = to;    e[tot].cap = 1;    e[tot].next = head[from];    head[from] = tot++;    e[tot].to = from;    e[tot].cap = 0;    e[tot].next = head[to];    head[to] = tot++;}void rev_bfs(){    memset(num, 0, sizeof(num));    memset(d, -1, sizeof(d));    d[sink] = 0;    num[0] = 1;    queue <int> q;    q.push(sink);    while(!q.empty()){        int u = q.front(); q.pop();        for(int i = head[u]; i != -1; i = e[i].next){            int v = e[i].to;            if(d[v] != -1) continue;            d[v] = d[u] + 1;            q.push(v);            num[d[v]]++;        }    }}int ISAP(){    memcpy(cur, head, sizeof(cur));    rev_bfs();    int flow = 0, u = pre[source] = source, i;    while(d[source] < nv){        if(u == sink){            int f = INF, neck;            for(i = source; i != sink; i = e[cur[i]].to)                if(f > e[cur[i]].cap){                    f = e[cur[i]].cap;                    neck = i;                }            for(i = source; i != sink; i = e[cur[i]].to){                e[cur[i]].cap -= f;                e[cur[i]^1].cap += f;            }            flow += f;            u = neck;  //记录应该回退的点,不直接回退到源点，小优化        }        for(i = cur[u]; i != -1; i = e[i].next)  //找最短路            if(d[e[i].to] + 1 == d[u] && e[i].cap) break;        if(i != -1){            cur[u] = i;  //记录当前弧            pre[e[i].to] = u; //记录上一个节点            u = e[i].to;        }        else{  //u到sink出现断层            if(0 == (--num[d[u]])) break; //GAP间隙优化，如果出现断层，可以知道一定不会再有增广路了            int mind = nv;            for(i = head[u]; i != -1; i = e[i].next)                if(e[i].cap && mind > d[e[i].to]){                    cur[u] = i;   //修改当前弧                    mind = d[e[i].to];                }            d[u] = mind+1;            num[d[u]]++;            u = pre[u];  //回退到上一个点        }    }    return flow;}bool judge(int x, int y){    if(x < 1 || x > s || y < 1 || y > a) return false;    return true;}void init(){    memset(head, -1, sizeof(head));    tot = 0;}void solve(){    init();    scanf("%d%d%d", &s, &a, &b);    for(int i = 1; i <= s; i++){        for(int j = 1; j <= a; j++){            int loc = (i-1)*a+j;            addedge(loc, loc+s*a);            if(judge(i+1, j)){                addedge(i*a+j+s*a, loc);                addedge(loc+s*a, i*a+j);            }            if(judge(i, j+1)){                addedge((i-1)*a+j+1+s*a, loc);                addedge(loc+s*a, (i-1)*a+j+1);            }            if(i == 1 || j == 1 || i == s || j == a){                addedge(loc+s*a, 2*s*a+1);            }        }    }    int x, y;    for(int i = 1; i <= b; i++){        scanf("%d%d", &x, &y);        int loc = (x-1)*a+y;        addedge(0, loc);    }    source = 0, sink = 2*s*a+1;    nv = sink+1;    printf(b == ISAP() ? "possible\n" : "not possible\n");}int main(){    int T;    scanf("%d", &T);    while(T--) solve();    return 0;}