UVA 515 King

题目链接：http://acm.hust.edu.cn/vjudge/problem/25023



题意：有n个数(a1~an)，有m个关系：给出区间[L,R]和k，∑ai < k 或者 ∑ai > k。然后判断是否存在这样一个序列。



思路：差分约束模板题，需要把公式变一下型，构建n+1个点Si(0<=i<=n)表示前i个数的前缀和一个源点。然后关系就变成了Si - Sj > k通过两边取反或+1的形式把关系不等式变形成Si-Sj<=k，然后直接套模板。




#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <utility>
using namespace std;

#define rep(i,j,k) for (int i=j;i<=k;i++)
#define Rrep(i,j,k) for (int i=j;i>=k;i--)

#define Clean(x,y) memset(x,y,sizeof(x))
#define LL long long
#define ULL unsigned long long
#define inf 0x7fffffff
#define mod 100000007
const int maxn = 1009;

struct node
{
int to,w,next;
}edge[maxn];
int head[maxn];
int tot = 0;
int n,m;
void addedge(int x,int y,int z)
{
tot++;
edge[tot].w = z , edge[tot].to = y , edge[tot].next = head[x];
head[x] = tot;
}
int dis[maxn];

bool spfa( int st )
{
bool vis[maxn];
int outque[maxn];
memset(vis,0,sizeof(vis));
memset(outque,0,sizeof(outque));
for( int i = 0; i <=n+1; i++ ) dis[i] = inf;//初始化距离
queue<int> q;
q.push(st) , vis[st] = true , dis[st] = 0;//初始状态
while( !q.empty() )
{
int u = q.front();
q.pop();
outque[u]++ , vis[u] = false;
if ( outque[u] > n+2 ) return false; //有负环 无最短路

for( int k = head[u]; k != -1; k = edge[k].next )
{
int v = edge[k].to;
if ( dis[v] > dis[u] + edge[k].w )
{
dis[v] = dis[u] + edge[k].w;
if( !vis[v] ) vis[v] = true , q.push( v );
}
}
}
return true;
}

char str[maxn];

void init()
{
int x,y,k;
cin>>m;
Clean(head,-1) , tot = 0;
rep(i,1,m)
{
scanf("%d %d %s %d",&x,&y,str,&k);
if ( str[0] == 'g' )
addedge(x+y,x-1,-k-1);
else
addedge(x-1,x+y,k-1);
}
rep(i,0,n) addedge(n+1,i,0);
if ( spfa(n+1) )
puts("lamentable kingdom");
else
puts("successful conspiracy");
}

int main()
{
while( cin>>n )
{
if ( !n ) break;
init();
}
return 0;
}