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题目链接：点击打开链接

题意：给定s*a的方格点,有b个坐标是有且仅有一个人的。

每个点只能被经过一次

能不能让所有人都移动到矩阵边缘。

拆点一下，建图还是挺明显的。。

太卡了提交半天没结果，贴一下代码改天再搞好了。。

//好吧1A了。。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
#define ll int
const int MAXN = 100010;//点数的最大值
const int MAXM = 400010;//边数的最大值
const int INF = 0x3f3f3f3f;
struct Edge
{
    int to,next,cap,flow;
}edge[MAXM];//注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],cur[MAXN];
void add(int u,int v,int w,int rw = 0)
{
    edge[tol].to = v; edge[tol].cap = w; edge[tol].flow = 0;
    edge[tol].next = head[u]; head[u] = tol++;
    edge[tol].to = u; edge[tol].cap = rw; edge[tol].flow = 0;
    edge[tol].next = head[v]; head[v] = tol++;
}
int Q[MAXN];
void BFS(int start,int end)
{
    memset(dep,-1,sizeof(dep));
    memset(gap,0,sizeof(gap));
    gap[0] = 1;
    int front = 0, rear = 0;
    dep[end] = 0;
    Q[rear++] = end;
    while(front != rear)
    {
        int u = Q[front++];
        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].to;
            if(dep[v] != -1)continue;
            Q[rear++] = v;
            dep[v] = dep[u] + 1;
            gap[dep[v]]++;
        }
    }
}
int S[MAXN];
int sap(int start,int end,int N)
{
    BFS(start,end);
    memcpy(cur,head,sizeof(head));
    int top = 0;
    int u = start;
    int ans = 0;
    while(dep[start] < N)
    {
        if(u == end)
        {
            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 = false;
        int v;
        for(int i = cur[u]; i != -1; 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 != -1; 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 != start)u = edge[S[--top]^1].to;
    }
    return ans;
}
void init(){ tol = 0; memset(head,-1,sizeof(head)); }

int n,m,b;
int Hash(int x,int y){return (x-1)*m+y;}
int Hash2(int x,int y){return (x-1)*m+y + n*m;}
bool inmap(int x,int y){return 1<=x&&x<=n&&1<=y&&y<=m;}
int step[4][2] = {0,1,0,-1,1,0,-1,0};
int main(){
	int T, u, v, i, j;	scanf("%d",&T);
	while(T--){
		init();
		scanf("%d %d %d",&n,&m,&b);
		int from = 0, to = Hash2(n,m)+1;
		for(i = 0; i < b; i++)	
		{
			int px, py;
			scanf("%d %d",&px,&py);
			add(from, Hash(px,py), 1);
		}
		for(i = 1; i <= n; i++)
		{
			for(j = 1; j <= m; j++)
			{
				add(Hash(i,j), Hash2(i,j), 1);
				for(int k = 0; k < 4; k++)
				{
					int nowx = i+step[k][0], nowy = j+step[k][1];
					if(!inmap(nowx,nowy))continue;
					add(Hash2(i,j), Hash(nowx,nowy), INF);
				}
			}
		}
		for(i = 1; i <= n; i++)
		{
			add(Hash2(i,1), to, INF);
			add(Hash2(i,m), to, INF);
		}
		for(i = 1; i <= m; i++)
		{
			add(Hash2(1,i), to, INF);
			add(Hash2(n,i), to, INF);
		}

		b == sap(from, to, to+1) ? puts("possible"):puts("not possible");
	}
	return 0;
}

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Uva 563 网络流

标签：c   class   blog   code   a   http   

原文地址：http://blog.csdn.net/acmmmm/article/details/27109625